First, I will prove that $f(q) \ge q$ for all $q \in \mathbb{Q}_{\ge 1}$. Denote $P(i,x,y)$ be the assertion that if $i=1$, $f(x)f(y) \ge f(xy)$, and if $i=2$, $f(x+y) \ge f(x)+f(y)$. By $P(1,a,1)$, we have $f(a)f(1) \ge f(a)$, so $f(1) \ge 1$. By $P(2,1,n)$, we have $f(n+1) \ge f(n)+f(1) \ge f(n)+1$, so by induction $f(n) \ge n$ for all positive integers $n$. By $P(1,\frac{p}{q},q)$, we have $f\left(\frac{p}{q}\right) \ge \frac{f(p)}{f(q)} > 0$, so $f\left(\frac{p}{q}\right)>0$, which implies that $f$ is strictly increasing. This gives us $f(x) \ge f(\lfloor x \rfloor) \ge \lfloor x \rfloor > x-1$. By $P(1,x,x^k)$, we have $f(x)^k \ge f(x^k) > x^k-1$, so $f(x) > \sqrt[k]{x^k-1}$ for all $k$, or $f(x) \ge x$ for all $x \ge 1$. Claim. If $m$ is a fixed point of $f$, then so are all rational numbers in $[1,m-1]$. Proof of Claim. We have, for $1\le x \le m-1$, $$m=f(m) \text{ (} m \text{ is given to be fixed point})$$$$ \ge f(x)+f(m-x) \text{ (by } P(2,x,m-x))$$$$ \ge x+m-x=m \text{ (by } f(x) \ge x \text{ for all } x \ge 1)$$Equality must hold everywhere, giving $f(x)=x$ for all rationals in $[1,m-1]$ as desired. Meanwhile, $a^n = f(a)^n \ge f(a^n) \ge a^n$, which gives that $a^n$ is a fixed point. Since the sequence $a^n$ is unbounded (note that $a>1$), we have $f(x)=x$ for all $x$ in $[1, \infty)$. Now we have for all $p, q \in \mathbb{N}$, by $P\left(1,\frac{p}{q},q\right)$ and $P\left(2,\frac{p}{q}, \frac{kp}{q}\right)$ for $1 \le k \le q-1$, we have $$f\left(\frac{p}{q}\right) \ge \frac{f(p)}{f(q)} = \frac{p}{q}$$and $$f(p) \ge qf\left(\frac{p}{q}\right) \implies f\left(\frac{p}{q}\right) \le \frac{p}{q}$$which gives us $f\left(\frac{p}{q}\right) = \frac{p}{q}$, so $f(x)=x$ for all $x \in \mathbb{Q}_{>0}$, as desired.

rkm0959 wrote: By $P(1,a,1)$, we have $f(a)f(1) \ge f(a)$ I would like to know how could you get $f(1)\ge 1$ from $f(a)f(1) \ge f(a)$.The image of $f$ can be on $R$, $f(a)$ could be negative or something. Maybe I had missed something,please tell me.

Well, I think rkm chose $a$ from the problem statement, satisfying $f(a)=a>1$...

"$f(x)f(1)\geq f(x)$, which means $f(x)\geq 0$. A" - What if $f(1) = 1$?

Note that $af(1)=f(a)f(1)\ge f(a)=a$, so $f(1)\ge 1$, whence $$af(n)=f(a)f(n)\ge f(na)\ge nf(a)=na\Rightarrow f(n)\ge n,\ \forall n\in \mathbb{N}$$If there is a positive rational $q=\dfrac{m}{n}$ such that $f(q)\le 0$, then $0\ge f(q)f(n)\ge f(m)\ge m$, contradiction. We have that $f(x)=f \left ( \lfloor x \rfloor +\{ x\} \right )\ge f \left ( \lfloor x \rfloor \right )+f(\{x\})> \lfloor x \rfloor>x-1 $. If there exists a positive rational $x>1$ such that $f(x)<x$, say $f(x)=x-\varepsilon$, we have that $(x-\varepsilon)^n=f(x)^n\ge f(x^n)>x^n-1$ which is a contradiction for $n$ big enough. Therefore, $f(x)\ge x,\ \forall x\in \mathbb{Q}_{\ge 1}$. Take now a positive rational $x$ and let $k$ big enough so that $a^kx>1$. We have then $a^kf(x)=f(a)^kf(x)\ge f(a^kx)\ge a^kx$, so $f(x)\ge x,\ \forall x\in \mathbb{Q}_{+}$. Observe that $a^k=f(a)^k\ge f(a^k)\ge a^k$, so $f(a^k)=a^k$ for all $k\in \mathbb{N}$. If there exists a positive rational $x$ such that $f(x)>x$, take $k$ big enough so that $a^k>x$. We would have then $$a^k=f\left ( (a^k-x) +x\right )\ge f(a^k-x)+f(x)>(a^k-x)+x=a^k$$which is a contradiction. Therefore, $f(x)=x$ for all positive rationals.

From (ii) we get that $f(nx)\ge nf(x)$ for every $n\in\mathbb{Z_+}$ $f(n)f(x)\ge f(nx)\ge nf(x)\Longrightarrow (f(n)-n)f(x)\ge 0$ for every $n\in\mathbb{Z_+}$ Subs $x=a$, then $(f(n)-n)a\ge 0$, then $f(n)\ge n$ for every $n\in\mathbb{Z_+}$ Suppose that $f(n_0)>n_0$ for some $n_0\in\mathbb{Z_+}$ $(f(n_0)-n_0)f(x)\ge 0\Longrightarrow f(x)\ge 0$ for every $x$ $f(x)\ge f(y) + f(x-y)\ge f(y)$ for every $x>y$, then $f$ is increasing Let $\epsilon = f(n_0)-n_0$. Let $m_0>\frac{n_0}{\epsilon}$, $m_0\in\mathbb{Z_+}$ Let $N\in\mathbb{Z_+}$ such that $a^N > m_0n_0$. Let $q=\left\lfloor\frac{a^N}{n_0}\right\rfloor$ $$q+1>\frac{a^N}{n_0} > m_0\Longrightarrow\ \ q\ge m_0 > \frac{n_0}{\epsilon}\Longrightarrow\ \ q\epsilon > n_0$$$$q\le\frac{a^N}{n_0}<q+1\Longrightarrow\ \ qn_0\le a^N<qn_0+n_0<q(n_0+\epsilon)=qf(n_0)$$Then $qf(n_0)\le f(qn_0)\le f(a^N)\le f(a)^N = a^N < qf(n_0)$. Contradiction! Then $f(n) = n$ for every $n\in\mathbb{Z_+}$. Then for every $m,n\in\mathbb{Z_+}$: $$m=f(m)\ge nf\left(\frac mn\right) = f(n)f\left(\frac mn\right)\ge f(m) = m\ \ \Longrightarrow\ \ \ f\left(\frac mn\right) = \frac mn$$

Very elegant FE IMO 2013/5 wrote: Let $\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\mathbb Q_{>0}\to\mathbb R$ be a function satisfying the following three conditions: (i) for all $x,y\in\mathbb Q_{>0}$, we have $f(x)f(y)\geq f(xy)$; (ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$; (iii) there exists a rational number $a>1$ such that $f(a)=a$. Prove that $f(x)=x$ for all $x\in\mathbb Q_{>0}$. Proposed by Bulgaria Firstly, we see that $f$ takes only positive values. Put $x=a, y=1$ in $(1)$ implies $af(1)=f(a)f(1) \ge f(a)=a$; so $f(1) \ge 1$. Apply $(2)$ repeatedly to see that $f(nx) \ge nf(x)$ for all $x \in \mathbb{Q}^{>0}$ and $n \in \mathbb{N}$. We see that $f(n) \ge n$ so we have $$f\left(\frac{r}{s}\right)f(s) \ge f(r).$$Hence, $f\left(\tfrac{r}{s}\right)>0$ for all rational numbers of the form $\tfrac{r}{s}$ for pairwise coprime $r, s$ positive integers. Claim: $f(x) \ge x$ for all $x \in \mathbb{Q}^{>0}$. Note that $(2)$ also yields that $f$ is an increasing function. Suppose $x_0$ is a positive rational for which $f(x_0)<x_0$. Hence, $$\lfloor x_0^n \rfloor \le f\left(\lfloor x_0^n \rfloor \right) \le f(x_0^n) \le f(x_0)^n$$for all $n \in \mathbb{N}$. For $n$ big, we obtain $f(x_0)^n<x_0^n-1$ which yields a contradiction! $\blacksquare$ Note that for any $x<a$ we have $$a=f(a) \ge f(x)+f(a-x) \ge x+(a-x),$$so $f(x)=x$. Club $f(a)=a$ to conclude that $f(x)=x$ for all $x \leqslant a$. For any $p \le a$ we have $$p^n \le f(p^n) \le f(p)^n=p^n,$$so $f(p^n)=p^n$ for all $n \geqslant 1$. For any positive rationals $r, s$ such that $r \le sa \le ra^2$ we have $$f(r) \le f(s)f\left(\frac{r}{s}\right)=r\left(\frac{f(s)}{s}\right)$$and $$f(s) \le f(r)f\left(\frac{s}{r}\right)=s\left(\frac{f(r)}{r}\right),$$so $$\frac{f(r)}{r}=\frac{f(s)}{s}.$$Fix $\ell>\tfrac{1}{a-1}$ to be an integer; $t>a$ be any rational number and $k$ be the maximal integer such that $$\left(1+\frac{1}{\ell}\right)^k \le \frac{t}{a},$$then $$\frac{t}{a} \le \left(1+\frac{1}{\ell}\right)^{k+1} \le ta.$$To finish, notice that $$\frac{f(t)}{t}=\frac{f\left(\left(1+\frac{1}{\ell}\right)^{k+1}\right)}{\left(1+\frac{1}{\ell}\right)^{k+1}}=1.$$Hence, $f$ is the identity function, just as we desired. $\blacksquare$

Let $Q(x,y)$ denote the assertion of $f(x)f(y)\geq f(xy)$ and $P(x,y)$ the assertion of $f(x+y)\geq f(x)+f(y)$. Doing an easy induction and using $Q(a,x)$ we have that $a^nf(x) \ge f(a^nx)$. $\text{ }$$\text{ }$$\text{ }$$\text{ }$$\text{ }$$(1)$ Doing an easy induction and using $P(a,ka)$ where $k$ is a positive integer we have that $f(ak) \ge ak$. $\text{ }$$\text{ }$$\text{ }$$\text{ }$$\text{ }$$(2)$

Using $(1)$ and Claim 3 we have that $f(a^n)=a^n$, so there are infinitely many and arbitrarily large $X$ such that $f(X)=X$, but plugging in $P(x,y)$ with $x+y=X$ we have that $X\ge f(x)+f(y) \ge x+y=X$ so $f\equiv\text{id}$, so we are done.

From (i), we have $a=f(a)f(1)\geq f(a)=1$ which implies that $f(1)\geq 1$. We want to show that $f(1)=1$. Suppose $f(1)=c>1$. Then $f(x)f(1)\geq f(x)$ implies $f(x)\geq0$. Then there exists sufficiently large integers $k$ such that $f(a^k) \ge f(\left \lfloor{a^k}\right \rfloor)+f(a^k-\left \lfloor{x}\right \rfloor )\geq f(\left \lfloor{a^k}\right \rfloor) \geq \left \lfloor{a^k}\right \rfloor+1 $, a contradiction since $f(a^k)\leq a^k$ for postive integers $k$ and this can be easily seen be appliying (i) repeatedly. So $f(1)=1$. For each $q$, we choose positive integer $m>1$ and $m>q$ such that $qm$ is an integer. Now, $f(m)f(q) \ge f(mq)$. But we have both $f(m)$ and $f(qm)\ge 1$, so similarly we can do poof by contradiction to show that $f(n)=n$ for all positive integers $n$. Now, by (ii), $f(m)\ge f(m-q)+f(q)$ but also $f(m)f(q)\ge f(mq)$ and $f(m)f(m-q)\ge f(m(m-q))$, so $f(q)\ge q$ and $f(m-q)\ge m-q$. Hence $f(q)=q$ and we're done.

$\text{Throughout the proof,}$ $n$, $m$, $p$ $\text{and}$ $q$ $\text{always denote positive integers and}$ $k$ $\text{could be any positive real number. The others are all rational numbers.}$ $\text{(1): By strong induction in (ii), we have}$ $f(an)\ge an$ $ \forall n$ $\text{(2): By (i), (iii) and (1) we know that}$ $f(a)f(n)=af(n)\ge f(an)\ge an$ $\text{, therefore}$ $f(n)\ge n$ $ \forall n$ $\text{(3): If}$ $\exists\lambda=\frac{m}{n}$ $\text{s.t.}$ $f(\lambda)\le 0$ $\text{then by (i)}$ $f(an)f(\lambda)\ge f(am),$ $\text{contradiction by (1)!}$ $\text{(4): By (ii) and (3) it's trivial that}$ $f$ $\text{is strictly increasing.}$ $\text{(5): By strong induction in (i) we can conclude that}$ $a^n\ge f(a^n)$ $ \forall n.$ $\text{(6): Again, by strong induction using (ii), we have}$ $f(nx)\ge nf(x)$ $ \forall n$ $\text{If}$ $\exists\alpha$ $\text{s.t.}$ $f(\alpha)=\alpha+k$ $\text{we can take}$ $a^n$ $\text{arbitrarily large because}$ $a>1$ $\text{and a number}$ $\beta$ $\text{s.t.}$ $\alpha+\beta=a^n,$ $\text{then, by (5) and (ii):}$ $a^n\ge f(a^n)\ge f(\alpha)+f(\beta)=\alpha+k+f(\beta)$ $\text{so}$ $f(\beta)<\beta.$ $\text{From (2) and (4),}$ $f(\beta)>\lfloor \beta \rfloor$ $\text{then because of the last line,}$ $k<1$. $\text{In other words, if the output of a number}$ $x$ $\text{is bigger than himself, it must be lower than}$ $x+1.$ $\text{We take}$ $f(\alpha)=\alpha+k$ $\text{and from (6)}$ $f(n \alpha)\ge n \alpha+nk,$ $\text{but}$ $nk$ $\text{can be arbitrarily large, contradiction! therefore:}$ $\text{(8):}$ $f(x)\le x$ $\text{(9):}$ $f(n)=n \forall x$ $\text{Let}$ $x=\frac{p}{q}.$ $\text{From (ii):}$ $f(\frac{p}{q})f(q)\ge f(p),$ $\text{by (9):}$ $f(\frac{p}{q})\ge \frac{p}{q}$ $\text{and from (8):}$ $f(x)=x$

Let the fact that $f(x)f(y) \ge f(xy)$ be denoted by $P(x,y)$ and $f(x+y) \ge f(x) + f(y)$ be denoted by $Q(x,y)$. We extend the function $f$ for our convenience and let $f(0) = 0$, but note that we can't use $0$ in the given inequalities. Claim 1: We have $f(1) \ge 1$. Proof: We take $P(a,1)$ to see that $f(a)f(1) \ge f(a)$. Since $f(a) = a > 0$, we can cancel out $f(a)$ from both sides to see $f(1) \ge 1$. $\blacksquare$ Claim 2: For all integers $k$, we have $f(k) \ge k$. Proof: We note that by repeatedly applying (ii), we have $f(k) \ge kf(1) \ge k$. $\blacksquare$ Claim 3: For all $x \in \mathbb{Q}_+$, we have $f(x) > 0$. Proof: Suppose there exists $x \in \mathbb{Q}_+$ with $f(x) \le 0$. Let $x = \frac{p}{q}$ where $p$ and $q$ are positive integers. Then, we take $P(q,x)$ to see that $f(q)f(x) \ge f(p) > 0$. But $f(q) > 0$, so $f(x) > 0$, which is a contradiction. $\blacksquare$ Claim 4: The function $f$ is strictly increasing. Proof: Consider any $x,y \in \mathbb{Q}_+$ with $x < y$. Taking $Q(x,y-x)$, we see that $f(y) \ge f(x) + f(y-x) > f(x)$. $\blacksquare$ Claim 5: For all $x > 1$, we have $f(x) \ge x$. Proof: We note that by repetitively applying (i), we have $f(x)^n \ge f(x^n)$ for all integers $n$. By Claim 4, we have that \[f(x)^n \ge f(x^n) \ge f\left(\left\lfloor x^n \right\rfloor\right) \ge \left\lfloor x^n \right\rfloor,\]so \[f(x) \ge \sqrt[n]{\left\lfloor x^n \right\rfloor}.\]Then, we have that \[f(x) \ge \lim_{n \rightarrow \infty}\sqrt[n]{\left\lfloor x^n \right\rfloor} = x,\]and we are done. $\blacksquare$ Claim 6: For all $x$, we have $f(x) \le x$. Suppose that $f(x) > x$ for some $x$. Then, we define $x_1 = x$, $x_2 = 2x$, and define $x_n = nx$ for general $n$. We note that because $f(x) > x$, we must have that $f(x) \ge \frac{kx}{k-1}$ for some integer $k$. Then, for $n \ge k-1$, we have that \[f(x_n) \ge nf(x) \ge \frac{nkx}{k-1} = nx + \frac{nx}{k-1} \ge nx + x = (n+1)x = x_{n+1}.\]Now, we define a new sequence $\{y_m\}_{m=1}^{\infty}$ so that $y_m = x_{m+k-1}$ for all $m$. This ensures that $f(y_m) \ge y_{m+1}$ for all $m$. Now, we define $r = \left\lfloor \log_a y_1\right\rfloor + 1$. For some $j$, we must have that $y_j > a^r$ and $y_{j-1} \le a^r$. Now, we note that $a^r = f(a)^r \ge f(a^r) \ge a^r$, with the last part following from Claim 5. Thus, $f(a^r) = a^r$. But we have that \[a^r < y_j \le f(y_{j-1}) \le f(a^r) = a^r,\]absurd. Thus, we are done. $\blacksquare$ We remark that Claim 6 implies that $f(x) = x$ for $x > 1$ because of Claim 5. Since $f(1) \ge 1$, Claim 6 also shows that $f(1) = 1$. Now, it suffices to show that $f(x) = x$ for $x < 1$. Suppose that there exists some $x < 1$ with $f(x) \ne x$. Let this $x$ be $\frac{p}{q}$ where $\gcd (p,q) = 1$. Then, we take $P(x,q)$ to see that $f(x)f(q) \ge f(p)$. Since $f(q) = q$ and $f(p) = p$, we see that $f(x) \ge x$. But then Claim 6 shows $f(x) = x$, contradiction. Thus, we see that $f(x) = x$ for $x > 1$, $x=1$, and $x < 1$ and we are done. $\blacksquare$

Sorry for very poorly written solution. I was writing it up as I was solving it. We have the following size lemma. Lemma: We have that $f(x)>0$ for all $x\in\mathbb{Q}_{>0}$ and $f(n)\ge n$ for all $n\in\mathbb{N}$. Proof: First we show that $f(x)\ne 0$ for all $x\in\mathbb{Q}_{>0}$. Indeed, suppose we had some $z$ such that $f(z)=0$. Then, plugging in $z$ and $a/z$ into (i), we get that $0\ge f(a)=a$, which is clearly a contradiction as $a>1$. Now, plugging in $(1,1)$ into (i), we get that $f(1)^2\ge f(1)$, or $f(1)\ge 1$. Iterating (ii) gives that $f(n)\ge n$ for all positive integers $n$, as desired. It suffices now to show that $f(p/q)>0$, where $p$ and $q$ are positive integers. To see this, note that $(p/q,q)$ on (i) implies that \[f(p/q)\ge\frac{f(p)}{f(q)},\]and since $f(p)$ and $f(q)$ are both positive, this implies that $f(p/q)$ is positive, as desired. $\blacksquare$ Now, iterating (i) shows that \[f(a^n)\le f(a)^n=a^n.\]But \[f(a^n)\ge f(\lfloor a^n\rfloor)+f(\{a^n\})> f(\lfloor a^n\rfloor)\ge\lfloor a^n\rfloor\]if $a$ is not integer, and $f(a^n)\ge a^n$ if $a$ is integer, so regardless, $f(a^n)\ge\lfloor a^n\rfloor$, so \[\lfloor a^n\rfloor\le f(a^n)\le a^n.\]In particular, this implies that \[a^{mn}-1<f(a^{mn})\le a^{mn}.\]From (i) we have $f(a^n)>\sqrt[m]{f(a^{mn})}$, so we have \[\sqrt[m]{a^{mn}-1}<f(a^n)\le a^n.\]By making $m$ very large, this implies that $f(a^n)=a^n$. Now, suppose we had some $x\in\mathbb{Q}_{>0}$ such that $f(x)=cx$ where $c>1$. Then, iterating (ii) implies that $f(Nx)\ge cNx$. But since $f$ is weakly increasing from (ii) and the lemma, this contradicts the fact that $f(a^n)=a^n$ for arbitrarily large $n$. Therefore, $f(x)\le x$. Now, suppose that we had a rational $x>1$ such that $f(x)=cx$ where $c<1$. Then, iterating (i) gives that $f(x^n)\le c^nx^n$. Again, combined with the fact that $f$ is increasing, this contradicts that $f(a^n)=a^n$ for arbitrarily large $n$. Therefore, we have $f(x)=x$ for all rationals $x>1$. We see that $2=f(2)\ge 2f(1)$, so $f(1)\le 1$, but we know $f(1)\ge 1$, so $f(1)=1$ as well. Then, (i) gives $f(1/x)\ge 1/f(x)$. But we know $f(x)\le x$, so we get $f(1/x)\ge 1/x$. We already know from before that $f(1/x)\le 1/x$, so we have $f(1/x)=1/x$ for all rationals $x>1$. This shows that $f(x)=x$ for all positive rationals $x$, completing the solution.

I am probably missing something, but i think we can extend this to $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ Lemma: There exist arbitrarily small $\varepsilon > 0$ such that $f(\varepsilon) \geq \varepsilon$ Assume the contrary, that for all $x \in (0, A)$ we have $f(x) < x$. Then use the first condition to get $f(x)f(\frac{a}{x}) \geq a \iff f(\frac{a}{x}) \geq \frac{a}{f(x)}>\frac{a}{x}$ for all $x \in (0, A)$. This implies $f(x) > x$ for all $x \in (B, \infty)$. But also from the first condition we have $f(a^n) \leq a^n$ which gets arbitrarily large, so we have a contradiction. Now assume there exists some $x$ such that $f(x) = x-\delta$ for some $\delta > 0$. Using the second condition and induction we get $f(nx) \geq nf(x)$ for positive integers $n$. Also from the second condition we have that $f$ is strictly increasing. Now we have $x-\delta = f(x) \geq nf(\frac{x}{n}) \geq nf(\varepsilon) \geq n\varepsilon \implies \frac{x-\delta}{n} \geq \varepsilon$ for some $\varepsilon \leq \frac{x}{n}$ that satisfies the condition in the above lemma. But look at the union of intervals $[\frac{x}{n}, \frac{x-\delta}{n})$ for all positive integers $n$. When $n$ gets big enough we will have $\frac{x}{x-\delta} > 1+\frac{1}{n} \iff \frac{x}{n+1}>\frac{x-\delta}{n}$, so the union of these intervals will contain an interval in the form $(0, c)$. Taking $\varepsilon \in (0, c)$ such that it satisfies $f(\varepsilon)\geq \varepsilon$ gives a contradiciton, so we get $f(x) \geq x$ for all $x \in \mathbb{R}^+$ Now we prove $f(x)=x$ on $(0,a]$. Suppose $f(b) > b$ for $b < a$, Then $f(a) \geq f(a-b) + f(b) > a$, a contradiction. Now assume $f(x)=x$ on $(0, A)$ for $A>1$. Then using the first condition we get $x \geq f(x)$ for all $x \in (0, A^2)$. This by induction gives $f(x)=x$ everywhere $\blacksquare$

By (i), $f(a)f(1)\ge f(a)$, which gives $f(1)\ge 1$. By (i) again, $f(x)f(1)\ge f(x)$ for all $x$, meaning $f(x)\ge 0$ for all $x$. Now note that for all $r$ where $f(r)\ge r$, we have $f(a)f(\frac{r}{a})\ge f(r)$ by (i), so $f(\frac{r}{a}\ge \frac{r}{a}$. Thus $f(r)\ge r$ implies $f(\frac{r}{a})\ge \frac{r}{a}$, meaning there exist arbitrarily small $z$ for which $f(z)\ge z$. Now I will prove $f(x)\ge x$ for all $x$. Given $x$, choose arbitrarily small $z$, and let $k$ be the largest integer such that $kz<x$. Then by (ii) we have $f(x)=kf(z)+f(x-kz)\ge kf(z)\ge kz\ge x-z$. Since $x-z$ is arbitrarily close to $x$, we have $f(x)=x$. Hence, $f(x)\ge x$ for all $x$. Finally, note that for all $k$, by (i) we have $a^k=f(a)^k\ge f(a^k)$, meaning $f(a^k)=a^k$ so there are arbitrarily large fixed points. But for each fixed point $m$, by (ii) we have $m=f(m)\ge f(y)+f(m-y)$ for $0<y<m$, meaning $f(y)=y$ since $f(x)\ge x$ for all $x$. This implies $f(x)=x$ for all $x$ below some fixed point, so we are done.

Solved with dantaxyz. Claim: $f$ is strictly increasing. Proof: We know $f(a)f(1) \ge f(a)$, so $f(1)\ge 1$ since $f(a)=a>1$. Now, $f(n)\ge n$ for integers $n$ by spamming the second condition. Let $x=p/q$ for integers $p,q$. Then $f(p/q)f(q)\ge f(p)$, so $f(p/q) \ge f(p)/f(q)$, which is positive since $f(p),f(q)>0$. So $f(x)>0$ for all $x$. Now, $f(x+y) \ge f(x)+f(y) > f(x)$, so $f$ is increasing. $\blacksquare$ Claim: $f(x)=x$ for $x>1$. Proof: Suppose $f(z)<z$ for some $z$. Let $f(z)=z/c$ for $c>1$. Since $f(z^n) \le f(z)^n$ by spamming the first condition, we get $f(z^n) \le z^n/c^n$. Since $c>1$, the issue is that this can get very small. We have \[ z^n/c^n \ge f(z^n) > f(a \lfloor z^n/a \rfloor) \ge a\lfloor z^n/a \rfloor \ge z^n-a. \]The second step follows from $f$ increasing, and the third from $f(ka) \ge kf(a)=ka$ for $k\in \mathbb{Z}$. If $z>1$, then we get a clear contradiction by making $n$ large enough. This proves $f(x) \ge x$ for all $x\ge 1$. To prove that $f(x)=x$ for $x\ge 1$, notice that $f(a^n) \le f(a)^n=a^n$ for $n\ge 1$, so actually $f(a^n)=a^n$. Therefore, there are arbitrarily large fixed points in $f$, and since $f$ is increasing, this forces $f(x)=x$ for all $x\ge 1$. $\blacksquare$ This also forces $f(x)\le x$ for $x<1$, since if there was an $x<1$ with $f(x)>x$, then since $f(x+42) \ge f(x)+f(42)$, we get $x+42 \ge f(x)+42$, i.e. $f(x) \le x$, contradiction. Now we show $f(x) \ge x$ for $x<1$. If $x<1$, then take some $y> 1/x$. Then $f(x)f(y) \ge f(xy)$, so $f(x)y\ge xy$, so $f(x)\ge x$. Now, $f(x)\ge x$ and $f(x)\le x$, so $f(x)=x$ even for $x<1$. So $f(x)=x$ for all $x$.

Let $M(x, y)$ denote the first multiplicative condition, $A(x, y)$ denote the additive condition. First of all, we claim that $f(1) \geq 1$. Consider $M(a, 1): \ af(1) \geq a$, implying the desired result. Now, induction with $A(1,1)$ and for general positive integer $n$, $A(n, 1)$ yields $f(n) \geq n$ over positive integers. Now, consider $M(p/q, q)$ for $p$ and $q$ integers. We get $f \left( \frac{p}{q} \right) \geq \frac{f(p)}{f(q)} > 0$, thus $f$ is positive everywhere. Now, notice that by repeated $M$, we get $f(x)^n \geq f(x^n)$. However, obviously $f(x^n) \geq f \left( \lfloor x^n \rfloor \right) > x^n - 1$, hence combining, we get \begin{align*} f(x)^n > x^n - 1. \end{align*}We can take the limit as $n$ approaches infinity to get $f(x) \geq x$; however, this relies on the fact that $x^n - 1$ is unbounded, thus only works for $x > 1$. Recall that we already showed $f(1) \geq 1$, though, so we know $f(x) \geq x$ for all $x \in [1, \infty )$. We will now obtain a bunch of fixed points. Notice that \begin{align*} a^n = f(a)^n \geq f(a^n) \geq a^n, \end{align*}hence $f(a^n) = a^n$ for each positive integer $n$. We get even more fixed points; observe that \begin{align*} a^n = f(a^n) \geq f(x) + f(a^n - x) \geq x + a^n - x = a^n. \end{align*}This forces $f(x) \geq x$ and $f(a^n - x) \geq a^n - x$ to be equalities, and so any $x > 1$ is fixed. Now, to kill $x \leq 1$, take some large $N$. Notice that \begin{align*} Nf(x) = f(x)f(N) \geq f(Nx) = Nx, \end{align*}hence $f(x) \geq x$. We need another bound: \begin{align*} x + N = f(x + N) \geq f(x) + f(N) \geq x + N, \end{align*}so we are done. YUH.

First, as $f(a) = a,$ we get $f(1)f(a) \ge f(a)$ so that $f(1) \ge 1$. Also, we can easily prove that $f(a^n) \le f(a)^n = a^n.$ Lemma 1: $f(x) > 0$ for all $x$. Proof: This is clearly true for all integers, so let $x = \frac{p}{q}$ be a rational number. Then, $f(q)f(x) \ge f(p)$ so that $f(x) \ge \frac{f(p)}{f(q)}$, as desired. Observe that $f$ is now strictly increasing, as $f(x + \epsilon) \ge f(x) + f(\epsilon) \ge f(x)$. Lemma 2: $f(2) = 2$. Proof: Assume for contradiction $f(2) = 2 + k$ for some positive rational number $k$, so that $f(2n) \ge 2n + nk$ for all positive integers $n$. Next, let $2j$ be the greatest even number less than $a^n$ for some large $n$ (We can make $a^n$ as large as we want). Thus, $$a^n \ge f(a^n) > f(2j) \ge jf(2) = 2j + jk,$$implying that $a^n - 2j \ge jk$. However, $a^n - 2j$ is between $0$ and $2$ while we can grow $jk$ to be as big as we want. This is the desired contradiction. Now that $f(2) = 2$, we see that $f(2n) \ge nf(2) = 2n$ for all positive integers $n$, while $f(2^n) \le f(2)^n = 2^n$ for all positive integers $n$. Thus $f(2^n) = 2^n$ for all positive integers $n$ so we can get arbitrarily large integer fixed points. Next, note that $f(2^n) \ge f(x) + f(y) \ge x + y$ where $x + y = 2^n$ and $x,y$ are integers, so that $f(x) = x$ and $f(y) = y$. Thus $f(n) = n$ for all integers $n$. Next, let $x = \frac{p}{q}$ be a rational number. Then $p = f(p) = f\left(q\cdot\frac{p}{q}\right) \ge q\cdot f\left(\frac{p}{q}\right)$. Thus $f(x) \le x$ for all rational numbers $x$. Also, $1 = \frac{p}{q} \cdot \frac{q}{p} \ge f\left(\frac{p}{q}\right)f\left(\frac{q}{p}\right) \ge f(1) = 1$, so equality must hold, and $f\left(\frac{p}{q}\right) = \frac{p}{q}$ as desired.

Notice $f(a^n) \leq a^n$ by the first condition. Notice $f(x)f(a) \geq f(xa) \geq xf(a)$ for all positive integers $x$ so $f(x) \geq x$ for all positive integers $x$ by the first and second condition. Notice by plugging in the denominator, $f(x) \geq 0$ for all $x$ by the first condition. This implies $f$ is monotonic increasing. Notice $f(x) \geq x - 1$ and $f(x) \leq ax$ for $x \geq 1$ by monotonicity. So since some point $b$ lies below $y = x$, then $b^n$ dies by the first condition. If some point $b$ lies above $y = x$, then some interval around $b$ has slope greater than some constant greater than $a$, so an infinite Minkowski sum will span all rational numbers greater than some constant proving they have slope greater than $1$ by the second condition, which a sufficiently large $a^n$ will contradict. Thus all points lie on $y = x$ and we are done.

Claim 1: $f(x) \geq 0$ for all $x \in \mathbb{Q}^+$ First, notice that $f(1)f(a) \geq f(a) \implies f(1) \geq 1$. So, $f(x)f(1) \geq f(x) \implies f(x) \geq 0$. $\square$ The second condition clearly implies that $f$ is a strictly increasing function. Now, notice that for every positive integer $n$ we have $f(n) \geq n \cdot f(1) \geq n$. Also notice that $f(x)^k \geq f(x^k)$ for all $x \in \mathbb{Q}^+$ and $k \in \mathbb{N}$. Now, consider the following chain of inequalities. \[ f(x)^k \geq f(x^k) \geq f(\lfloor x^k \rfloor) \geq \lfloor x^k \rfloor \]So, $f(x) \geq \sqrt[k]{\lfloor x^k \rfloor}$. But, we have \[ \lim_{k \to \infty} \sqrt[k]{\lfloor x^k \rfloor} = x \implies f(x) \geq x \]But, notice that $a^n = f(a)^n \geq f(a^n)$ and $f(a^n) \geq a^n$. So, $f(a^n)=a^n$ for all positive integers $n$. Thus, $f$ has arbitrarily large fixed points. Which should just imply the result pretty easily (sorry I'm tired).

Claim $$f(m)\geq m \forall m\in\mathbb{N}$$Proof Notice how for $m$ natural we get $$f(m)f(y)\geq f(my)\geq mf(y)$$ Where the second inequlity follows by trivially inducting on $m$ .Now as for $y=a$ we get $f(a)$ positive, then $f(m)\geq m$ for all $m$ natural. \hfill{\blacksquare} Claim $f(q)>0$ for all $q\in\mathbb{Q}_{>0}$ Proof If $q$ is a natural number we're done. Thus let $q=\frac{k}{l}$. Now $$ f(q)f(l)\geq f (k) \geq k>0$$And as $f(l)\geq l$ then this implies $f(q)>0$. \hfill{\blacksquare} Claim $f(m)=m$ for all $m\in\mathbb{N}$ Proof. Notice how by inducrion on $n$ we obtain the following $$f(a^n)\leq a^n$$Thus now for any natural $m$ we have $$a^n\geq f(a^n)\geq f\left(m\lfloor \frac{a^n}{m}\rfloor\right) + f\left(a^n -m\lfloor \frac{a^n}{m}\rfloor \right)\geq f(m)\lfloor \frac{a^n}{m}\rfloor + f\left(a^n -m\lfloor \frac{a^n}{m}\rfloor\right)\geq f(m)\lfloor \frac{a^n}{m}\rfloor$$ But now if we let $f(m)=m+\delta$ we get that $$RHS=m\lfloor \frac{a^n}{m}\rfloor +\delta \lfloor \frac{a^n}{m}\rfloor \geq a^n-m+\delta \lfloor \frac{a^n}{m}\rfloor$$ But as $n\to +\infty$ we get $\lfloor \frac{a^n}{m}\rfloor > \frac{m}{\delta}$, so the inequality ends up being false unless $\delta =0$. \hfill{\blacksquare} Now we end by $$1=f(1)\geq bf\left(\frac{1}{b}\right)=f(b)f\left(\frac{1}{b}\right)\geq 1\implies f\left(\frac{1}{b}\right)=\frac{1}{b}$$ But also $f\left(\frac{a}{b}\right)\geq \frac{a}{b}=f(a)f\left(\frac{1}{b}\right)\geq f\left(\frac{a}{b}\right)\implies f\left(\frac{a}{b}\right)=\frac{a}{b}$

Solved with 20% Hint on ARCH

Let $P(x;y)$ be the first assertion and $Q(x;y)$ the second one. $P(1;a) \Rightarrow f(1) \ge 1$ and applying $Q$ multiple times we get $f(n) \ge nf(1) \ge n$, for any natural $n$. Also the function is increasing, because $f(\frac{a}{b}) \ge \frac{f(a)}{f(b)} >0$, so $f(x+y) \ge f(x) + f(y) >f(x)$ (Just basic observations untill now). Now assume $f(n)>n$ for some natural $n$, say $f(n)=n+a$, now we know $f(mn) \ge mf(n)= mn + ma$, and picking $m > \frac{1}{a}$, we get $f(x) > x+1$ for a natural $x$ (and as $f(k) \ge f(k-1) +f(1) > k+1$, $f(k) > k+1$ for $k \ge x$), meaning, for $q>x$ we have $f(q) > f(\lfloor{q}\rfloor) > \lfloor{q}\rfloor +1 > q $, which is contradicted by the fact than $f(a^n) \le f(a)^n=a^n$. So now we know $f(n)=n$, this means that $nf(\frac{1}{n}) \ge 1$ (from $P( n; \frac{1}{n})$) and $nf(1/n) \le f(1)=1$ (from $Q$), now $f(\frac{1}{n})=\frac{1}{n}$, which leads to: $\frac{n}{m}=f(\frac{1}{m})f(n) \ge f(\frac{n}{m})$ and also $f(\frac{n}{m})f(m) \ge f(n) \Rightarrow f(\frac{n}{m}) \ge \frac{n}{m}$, thus $q \ge f(q) \ge q$ for any rational $q$ and this completes the proof.

Note that $$(a,1)\mapsto f(a)f(1)\ge f(a)\stackrel{f(a)\ne0}{\Rightarrow}f(1)\ge1\Rightarrow f(n)\ge nf(1)\ge n\stackrel{(p/q,q)}{\Rightarrow}f(\frac pq)\ge\frac{f(p)}{f(q)}>0\Rightarrow\forall q_1>q_2\in\mathbb{Q}:f(r_1)\ge f(r_2)+f(r_1-r_2)>f(r_2)$$$$\implies f(x)\ge f(\lfloor x \rfloor)\ge\lfloor x\rfloor > x-1\Rightarrow f(x)=\sqrt[k]{f(x)^k}\ge\sqrt[k]{f(x^k)}\ge\sqrt[k]{f(\lfloor x^k\rfloor)}\ge\sqrt[k]{\lfloor x^k\rfloor}\implies f(x)\ge x$$for infinitely large k (this works, right?). Then, $$a^n=f(a)^n\ge f(a^n)\ge a^n\implies a^n=f(a^n)\ge f(x)+f(a^n-x)\ge(x)+(a^n-x)=a^n\implies f(x)=x\forall x<a^n,$$and since we can get n arbitrarily large by $a^n=f(a^n)$, we conclude $f(x)=x\forall x\in\mathbb Q_{>0}$.

Set $y=1$ in (i) to get $f(x)(f(1)-1) \ge 0$ so either $f(1) \ge 1$ and $f(x) \ge 0$ for all $x$ or $f(1)\le 1$ and $f(x) \le 0,$ contradiction because of (iii). Thus $f(x) \ge 0.$ Now suppose $f(k)=0$ for some $k.$ Setting $y=k$ gives $f(xk)=0$ for all $x,$ contradiction because of (iii). Thus $f(x)>0$ so from (ii) $f$ is increasing. Furthermore from $f(1) \ge 1$ and (ii) we easily get $f(n) \ge n$ for integers $n,$ so $f(x) \ge \lfloor x \rfloor$ for all $x.$ Now consider any $k$ and suppose $f(k)=k-c$ for some positive $c.$ From (i) we get $f(k)^n \ge f(k^n) \ge \lfloor k^n \rfloor \ge k^n-1.$ However, for large enough $n$ we have $(k-c)^n+1<k^n$ so this is impossible so we get $f(k) \ge k.$ Now set $x=y=a$ in (i) to get $f(a^2)=a^2$ and similarly by induction $f(a^n)=a^n$ for any $n.$ Now in (ii) setting $y=a^n-x$ gives $f(x)=x$ for all $x<a^n.$ By making $n$ arbitrarily large, we are done.

Our solution follows with two main claims: Claim 1: $f$ only takes positive values and is increasing. First note that $f(1)f(a) \ge f(a) \implies f(1) \ge 1$, and a quick induction using the second condition tells us $f(m) \ge m$ for $m \in \mathbb{Z}^+$. Thus we know $f \left(\frac mn\right) \ge \frac{f(m)}{f(n)} > 0$ for all rationals $\frac mn$, and the increasing behavior follows through the second condition. ${\color{blue} \Box}$ Claim 2: $f(x)=x$ for all $x \in \mathbb{Q}_{>0}$. $x>1: \quad$ Consider a positive integer $k$. Then \[f(x)^k \ge f(x^k) \ge f(\lfloor x^k \rfloor) > x^k-1,\] so if $f(p)<p$ for some $p$, we can make $k$ sufficiently large to get a contradiction. Thus $f(x) \ge x$ for all $x$. We go further by using our fixed point $a$ by noting \[a^k \ge f(a^k) \ge f(x)+f(a^k-x) \ge a^k \implies f(x)=x.\] $x=1: \quad 2 = f(2) \ge 2f(1) \ge 2$, so $f(1)=1$. $x<1: \quad$ Take a sufficiently large integer $k$ such that $kx \ge 1$. Then \[kx = f(kx) \ge kf(x) = f(k) f(x) \ge f(kx) = kx \implies f(x) = x. \quad \blacksquare\]

Let $n \geq 2$ be a positive integer:\[af(n) = f(a)f(n) \geq f(na) = f(a) + f(a) + \ldots + f(a) \geq nf(a) \Rightarrow f(n) \geq n.\]If $n = 1$, then we can actually finish at the step $af(1) = f(a)f(1) \geq f(a) = a \implies f(1) \geq 1$, hence $f(x) \geq x$ for all $\mathbb{Z}_{> 0}$. This also tells us $f$ is always positive since $f(1)f(x) \geq f(x)$ always. Let's try extending this to rationals. First of all, for any rational $r > 1$, note $f(r) \geq f(\lfloor r \rfloor) \geq x - 1$. Next we can show $f(x) \geq x$ for $x \in \mathbb{Q}_{> 1}$. Suppose there is a $r$ for which this is not the case, where $f(r) < r$. Then, $f(r)^n \geq f(r^n) \geq r^n - 1$ which is clearly untrue for large enough $n$. Then, note, for any powers of $a$, that $a^n = f(a)^n \geq f(a^n) \geq a^n$ since $a^n > 1$ always, hence $f(a)^n = f(a^n) = a^n$. Even when rational $x < 1$ we can consider $a^nx > 1$ and\[a^nf(x) = f(a)^nf(x) \geq f(a^nx) \geq a^nx\]since $a^nx > 1$ so indeed $f(x) \geq x$ for all positive rationals $x$. Now say there exists $r$ with $f(r) > r$; then take some $c$ power of $a$ large enough greater than $r$ so that\[f(c) \geq f(c - r) + f(r) > c\]which is bad since $f(c) = c$ for any $c = a^n$, contradiction. We are done. $$\mathbb{Q.E.D.}$$

We begin by showing that $f(x)>0$. Substituting $x=1, \ y=a$ into $(i)$, we get $f(1)f(a)=f(a) \implies f(1) \geq 1 > 0$. Then, by condition $(ii)$, we have $f(x) \geq xf(1) > 0$ for all $x \in \mathbb{N}$. Substituting $y=\frac{1}{x}$, we have $f(x)f(\frac{1}{x}) \geq f(1) > 0 \implies$ $f(\frac{1}{x})>0$ for all $x \in \mathbb{N}$. Then, by condition $(ii)$, $f(\frac{y}{x}) \geq yf(\frac{1}{x}) > 0$ for all $x, y \in \mathbb{N}$. $\square$ Now, we show that $x \geq f(x)$. Suppose FTSOC that some $x\in \mathbb{Q}_{>0}$ satisfies $f(x)>x$. Let $f(x)=x+ \epsilon$. Consider a sufficiently large positive integer $k$. Applying a simple induction to condition $(i)$ yields us $a^k \geq f(a^k)$. Let $p$ be the largest positive integer such that $a^k \geq px$. Note that $p > \frac{a^k}{x}-1$ Now, applying condition $(ii)$ repeatedly, we get $f(a^k) \geq f(px)+ f(a^k-px)>f(px)$. Then by using induction, $a^k \geq f(a^k) \geq f(px) \geq pf(x) = p(x + \epsilon) > a^k-x + \epsilon(\frac{a^k}{x}-1)$. Hence, we must have $x> \epsilon(\frac{a^k}{x}-1)$ for all sufficiently large integers $k$. This cannot be true unless $\epsilon=0$. $\square$ The rest is smooth sailing. Substituting $y=\frac{a}{x}$, we obtain $\frac{a}{x}f(x) \geq f(\frac{a}{x})f(x) \geq f(a)=a \implies f(x) \geq x \implies f(x)=x$. $\blacksquare$

Let $P(x,y)$ denote the first assertion and let $Q(x,y)$ denote the second assertion. Let $R(x,k)$ denote the assertion that $f(kx) \geq kf(x)$ for integers $k$, which follows from repeatedly applying $Q$. Claim: We have $f(k ) \geq k$ for all positive integers $k$. Proof: We prove this with induction on $k$: For $k=1$, this is true since $P(a,1)$ gives us $f(a)f(1) \geq f(a)$, and cancelling out $f(a)=a$ gives us $f(1) \geq 1$. For $k > 1$, taking $Q(k-1,1)$ gives us $f(k) \geq f(k-1) + f(1) \geq k$. Claim: We have $f(x) \geq 0$ for all $x$. Proof: We split into cases: Suppose $f(k) = k$ for all integers $k$. Then, for all pairs of positive integers $m$ and $n$, taking $P(\tfrac{m}{n}, n)$ gives us $f(\tfrac{m}{n}) \geq \tfrac{f(m)}{f(n)} = \tfrac{m}{n}$. Meanwhile, $R(\tfrac{m}{n}, n)$ gives us $f( \tfrac{m}{n} \cdot n ) \geq n f( \tfrac{m}{n} )$, so $f( \tfrac{m}{n} ) \leq \tfrac{m}{n}$. So, $f( \tfrac{m}{n} ) = \tfrac{m}{n}$ for all $m$ and $n$. This proves the whole problem, which certainly implies the claim. Suppose there exists an integer $k$ for which $f(k) > k$. Then, combining $P(x,k)$ with $R(x,k)$ gives us \[f(k)f(x) \geq f(kx) \geq kf(x) \implies f(x) (f(k) - k) \geq 0 \implies f(x) \geq 0.\] It follows from the second condition that $f$ is weakly increasing. Combining this with our first claim, it follows that $f(x) \geq \lfloor x \rfloor$ for all $x$. Claim: We have $f(x) \geq x$ for all $x \geq 1$. Proof: Suppose FTSOC that $j > 1$ is such that $f(j) < j$. Then, applying $P(j^{k-1}, j)$ gives us \begin{align*} f(j) f(j^{k-1} ) &\geq f(j^k) \\ j \left( j^{k-1} - f(j^{k-1}) \right) < j^k - f(j) f(j^{k-1} )& \leq j^k - f(j^k) \\ j^{k-1} (j - f(j)) & \leq j^k - f(j^k), \end{align*}contradicting the fact that $f(j^k) \geq \lfloor j^k \rfloor$ for sufficiently large $k$. Claim: We have $f(a^k) = a^k$ for all $k \geq 1$. Proof: It suffices to show that $f(a^k) \leq a^k$, which follows from induction on $k$ and taking $P(a^{k-1}, a)$. Claim: We have $f(x) \geq x$ for all $x$. Proof: Let $k$ be such that $xa^k > 1$. Then the claim follows since \[f(x) a^k = f(x)f(a^k) \geq f(xa^k) \geq xa^k.\] Claim: We have $f(x) = x$ for all $x$. Proof: Let $k$ be sufficiently large, so that we have \[a^k = f( a^k) \geq f(x) + f(a^k - x) \geq f(x) + a^k - x,\]so $f(x) \leq x$. Combining this with $f(x) \geq x$ finishes.

Claim: For any integer $m$, $f(m) \ge m$. Proof: $$f(1)f(a) \ge g(a) \implies af(1) \ge a \implies f(1) \ge 1$$$$ f(m) \ge \underbrace{f(1)+ f(1)+ \cdots + f(1)}_{m \text{ times}} \ge m.$$ Claim: $f(x)>0$ for all $x$. Proof: Suppose for the sake of contradiction for positive integers $m$, $n$. $f(\tfrac{m}{n}) <0$ observe that $$0> f(\tfrac{m}{n}) \ge \underbrace{f(\tfrac1{n}) + f(\tfrac1{n})+ \cdots + f(\tfrac1{n})}_{m \text{ times}} \ge mf(\tfrac{1}{n}) \implies 0> \tfrac{1}{n}$$Now note that $f(\tfrac{1}{n})f(n) \ge f(1) \ge 1$ however since $f(n) \ge n >0$ then $f(\tfrac{1}{n})>0$ a contradiction. Claim: $f$ is increasing. Proof: $$f(x+y) \ge f(x)+f(y) > f(x).$$ Claim: For $x>1$, $f(x) \ge x$ Proof: $$f(x)^m \ge f(x^m) \ge \lfloor x^m \rfloor + f(\{ x^m \}) > x^m-1$$$$f(x) > \sqrt[m]{x^m-1}$$$$f(x) \ge \lim_{m \rightarrow \infty} \sqrt[m]{x^m-1} = x.$$ Claim:$f(a^d) =a^d$ for all positive integers $d$. Proof: $$a^d=f(a)^d \ge f(a^d) \ge a^d \implies f(a^d)=a^d.$$ Claim: $f(\tfrac{a^d}{m}) = \tfrac{a^d}{m}$ for all positive integers $d$, $m$ such that $\tfrac{a^d}{m} \ge 1$. Proof: $$a^m = f(a^m) \ge df(\tfrac{a^m}{d}) \ge a^m \implies f(\tfrac{a^m}{d}) = \tfrac{a^m}{d}.$$ Claim: $f(x)=x$ for all $x>1$. Proof: For each $x$ we can choose $\tfrac{a^m}{d}$ infinitely close to $x$ on both sides. This combined with $f$ increasing forces $f(x)=x$ for $x>1$. To finish suppose we have integers $m>n \ge 1$ now observe $$f(\tfrac{m}{n}) f(\tfrac{n}{m}) \ge f(1) \ge 1 \implies f(\tfrac{n}{m}) \ge \tfrac{n}{m}$$$$n=f(n) \ge mf(\tfrac{n}{m}) \ge n \implies f(\tfrac{n}{m}) = \tfrac{n}{m}$$$$1=f(\tfrac{m}{n}) f(\tfrac{m}{n}) \ge f(1) \ge 1 \implies f(1)=1$$Therefore $f(x)=x$ for all $x$ and we are done. $\blacksquare$

My favourite of all! Claim: For $n \in \mathbb N$ \[f(n)\geq n\] If $f(ka) \geq ka$ then \[f(ka+a)\geq f(ka)+f(a)\geq ka+a = (k+1)a\]Therefore by induction we have $f(ka)\geq ka$ for all $k \in \mathbb N$. Now by $(i)$ we get \[f(a)f(k)\geq f(ak)\geq ak\]Therefore $f(k)\geq k$ for $k\in \mathbb N$. Claim: \[f(n)=n\]for $n \in \mathbb N$ We prove $f(1)=1$. Using $(i)$ if we have $a^{k-1} \geq f(a^{k-1})$ we can get \[ a^{k} \geq f(a^{k-1}) \cdot f(a) \geq f(a^k)\]Hence by induction we have $a^k \geq f(a^{k})$ for $k \in \mathbb N$. Also we can have $f(k) \geq k\cdot f(1)$ by $(ii)$. Observe $f(a^k) \geq f(\lfloor a^k \rfloor)$ true by $(ii)$ Hence \[a^k \geq f(a^k) \geq f(\lfloor a^k \rfloor) \geq \lfloor a^k \rfloor \cdot f(1) \]This give us $\frac{a^k}{\lfloor a^k \rfloor} \geq f(1)$. As $f(1)\geq 1$ we have \[1+\frac{\{a^k\}}{\lfloor a^k \rfloor} \geq 1 + (f(1)-1)\]Observe that $\{a^k\}$ will decrease and $\lfloor a^k \rfloor$ will increase as $k$ grow. Hence we will get $f(1)-1 > \frac{\{a^k\}}{\lfloor a^k \rfloor}$ for sufficiently large $k$. Hence we get $f(1)=1$. Now we will prove $f(n)=n$ in same way. First of all notice $f(nk) \geq kf(n)$ by $(ii)$. Assume $n | \lfloor a^k \rfloor - g$ for $0 \leq g < n$. So we get \[a^k \geq f(a^k) \geq f(\lfloor a^k \rfloor) \geq f(\lfloor a^k \rfloor - g) \geq \frac{\lfloor a^k \rfloor - g}{n}f(n)\]Hence \[n \cdot \frac{a^k}{\lfloor a^k \rfloor - g} \geq f(n)\]For sufficiently large $k$ this won't hold true except $f(n)=n$. Claim: \[f\left( \frac{a}{b}\right)=\frac{a}{b}\]for $a,b >0$. \[f\left(\frac{1}{n}+\frac{n-1}{n}\right) \geq f\left(\frac{1}{n}\right) + f\left(\frac{n-1}{n}\right) \geq 2\cdot f\left(\frac{1}{n}\right) + f\left(\frac{n-2}{n}\right) \geq \cdots \cdots \geq n\cdot f\left(\frac{1}{n}\right)\]Hence $\frac{1}{n}\geq f\left(\frac{1}{n}\right)$. But \[f\left(n\right)f\left(\frac{1}{n}\right) \geq f(1) \Rightarrow f\left(\frac{1}{n}\right) \geq \frac{1}{n} \]give us $f\left(\frac{1}{n}\right)=\frac{1}{n}$. Using same argument we get \[f\left(\frac{a}{b}\right) \geq a \cdot f\left(\frac{1}{b}\right) \Rightarrow f\left(\frac{a}{b}\right) \geq \frac{a}{b} \]But again using $(i)$ we get \[f(a)f\left(\frac{1}{b}\right) \geq f\left(\frac{a}{b}\right) \Rightarrow \frac{a}{b} \geq f\left(\frac{a}{b}\right)\]Hence we get $f\left(\frac{a}{b}\right)=\frac{a}{b}$. Therefore $f(x)=x$ for all positive rational numbers $x$.

More solutions. We first have the following claim. Claim : $f(k)\geq k$ for all positive integers } $k$ Proof : Notice that $$af(k)=f(a)f(k)\geq f(ak)\geq kf(a)=ka$$Thus, $af(k)\geq ak\implies f(k)\geq k$ Now, consider a rational number $x$. Notice that $$f(x)^n \geq f(x^n) \geq f(\lfloor x^n \rfloor)\geq \lfloor x^n \rfloor \implies f(x)\geq \sqrt[n]{\lfloor x^n \rfloor}$$But this quantity approaches $x$ for adequately large $n$. Thus, $f(x)\geq x$ for all rationals $x>1$. Next, consider the following claim. Claim : $f(a^n)=a^n$ for all $n \in \mathbb{N}$ Proof : First, notice that $$f(a^n) \leq f(a)^n = a^n$$Further, since $a>1$, $a^n>1$ as well. Thus, by our previous observation, $f(a^n)\geq a^n$, which gives $$a^n \geq f(a^n)\geq a^n$$Thus, we must have $f(a^n)=a^n$ for all $n \in \mathbb{N}$. Now, consider a positive integer $k$. Select a positive integer $q$ such that $a^q -k >1$ (such $q$ exists as since $a>1$, $a^n$ is strictly increasing as $n$ increases). Then, $$a^q=f(a^q) \geq f(k)+ f(a^q-k)\geq k + a^q -k = a^q$$Thus, we must have $f(k)+ f(a^q-k)=a^q$. But, $f(k)\geq k$ and $f(a^q-k)\geq a^q-k$, so inorder for equality to hold we must have $f(k)=k$ and $f(a^q-k)=a^q-k$. Thus, for all positive integers $k$, $f(k)=k$. We now continue to extend this to the rationals. Consider a rational number $x=\frac{p}{q}$. Notice now that, $$f(\frac{p}{q})\geq \frac{f(p)}{f(q)}=\frac{p}{q}$$and also $$qf(\frac{p}{q})\leq f(p) \implies f(\frac{p}{q})\leq \frac{f(p)}{q}=\frac{p}{q}$$which in turn gives us that $$x\geq f(x) \geq x$$and thus $f(x)=x$ for all rational numbers $x>0$.

cool! Claim: $f(n) \geq n$ for all $n\in\mathbb N$ Proof: $P(a, 1)\colon f(1)f(a) \geq f(a) \implies (f(1) - 1)f(a)\geq 0,$ but since $a \geq 0, f(1)\geq 1.$ Using property 2, we are done. Claim:$f(x)\geq x$ for all $x\in \mathbb Q_{\geq1}.$ $f$ is also strictly increasing over all positive rationals. Proof: Let $k = \frac mn$ where $m, n \in \mathbb N, (m, n) = 1$. Then, $P(k, n)\colon f(k) \geq \frac{f(m)}{f(n)} > 0.$ By the 2nd condition, we have that $f$ is strictly increasing. Hence, $f(x) \geq f(\lfloor x\rfloor) \geq \lfloor x \rfloor > x - 1.$ Let $k\in\mathbb N,$ so by $P(x, x^k),$ we have $f(x)^k\geq f(x^k) > x^k - 1.$ Thus, $f(x)> \sqrt[k]{x^k - 1}.$ Thus, it follows that $f(x) \geq x.$ Claim: If $f(m) = m$ for some $m > 1,$ then $f(x) = x$ for all rationals $x$ in $[m - 1, 1].$ Proof: $m = f(m) \geq f(x) + f(m - x) \geq x + m - x = m$ where equality holds if and only if $f(x) = x, f(m - x) = m - x.$ Now, we can finish. Claim: $f(x) = x$ for all $x\in\mathbb Q_{\geq1}.$ Proof: We prove that $f(a^k) = a^k$. Simply note that $a^k = f(a)^k \geq f(a^k) \geq a^k.$ Thus, by applying the previous claim, this is done. Claim: $f(x) = x$ for all $x\in\mathbb Q_{>0}$ Proof: Let $\mathbb Q_{ > 0}\ni k = \frac mn$ for $m, n\in\mathbb N.$ Then, $m = f(m)\geq nf(\frac mn) \geq n \frac mn = m.$ $\blacksquare$