tenplusten wrote: Find all functions $f:\mathbb{R^+}\to\mathbb {R^+} $ such that for all $x,y\in R^+$ the followings hold: $i) $ $f (x+y)\ge f (x)+y $ $ii) $ $f (f (x))\le x $ i) implies $f(x)$ is strictly increasing ii) implies then $f(x)\le x$ i) implies $f(x)\ge f(x-y)+y>y$ $\forall y\in(0,x)$ Setting there $y\to x^-$, we get $f(x)\ge x$ $\forall x$ And so $\boxed{f(x)=x\quad\forall x>0}$ which indeed is a solution.

I got the first 3 steps ,can you please eloborate what you do in your final step?

tenplusten wrote: I got the first 3 steps ,can you please eloborate what you do in your final step? We have $f(x)>y$ $\forall y\in(0,x)$ Set there $y$ as near as you want of $x$ (but below) and this gives $f(x)\ge x$

another easy solution x≥f(f(x))≥f(f(x)-k)+k if k≥x we are done so x≥f(x) if there is at least one x wich x-s≥f(x) where s≠0 so we write x-s≥f(x)≥f(x-y)+y now we put x-s=y so that means f(x)=x

Clearly $f(x+y) \geq f(x) + y > f(x)$, i.e. $f$ is strictly increasing. If we suppose that $f(x_0) > x_0$ for some $x_0$, then the monotonicity and the second condition yield $x_0 \geq f(f(x_0)) > f(x_0) > x_0$, contradiction. Hence $f(x) \leq x$ for all $x$. Suppose $f(x_0) < x_0$ for some $x_0$. Then in the first condition replace $x$ with $x_0-f(x_0)$ and $y$ with $f(x_0)$ to obtain $f(x_0) \geq f(x_0 - f(x_0)) + f(x_0) > f(x_0)$, impossible. Hence $f(x) \geq x$ for all $x$. In conclusion, $f(x) = x$ for all $x$. Direct verification shows that this function is indeed a solution.

Clearly $f$ is strictly increasing and so $f(x)\leqslant x$. If $f(x)<x$ for some $x$, then $(x,y)=(x-f(x),f(x))$ gives contradiction. So $f\equiv \text{Id}$, which works.

Let $P(x,y)$ the assertion of the F.E. (ineq actually). Clearly as $f(x+y) \ge f(x)+y>f(x)$ we have $f$ strictly increasing. Suppose there exists $x$ with $f(x)>x$, then by $P(x,f(x)-x)$ we get $x \ge f(f(x)) \ge 2f(x)-x$ which implies $x \ge f(x)$ contradiction!. Therefore $x \ge f(x)$ for all positive reals $x$, which also means that as $x$ becomes smaller so does $f(x)$, so now lets consider $P(x-y,y)$ for $x>y$ this gives $f(x) \ge f(x-y)+y$, if we assume FTSOC $x>f(x)$ for some $x$ then this gives $0 \ge f(x-f(x))$, contradiction!. Therefore $x \ge f(x) \ge x$ so $f(x)=x$ is the only solution, thus we are done .

Pretty straightforward. The answer is $f(x) = x$ for all $x \in \mathbb{R}^+$. It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions. Let $P(x,y)$ denote the assertion that $f (x+y)\ge f (x)+y $ for positive reals $x$ and $y$. We first show the following property of $f$. Claim : The function $f$ is (strictly) increasing. Proof : For any positive $\epsilon$ , $P(x,\epsilon)$ yield, \[f(x+\epsilon)/ge f(x)+\epsilon > f(x)\]which implies the claim. Say there exists some positive real $x_0$ such that $f(x_0)>x_0$. Then, $P((f(x_0)-x_0),x_0)$ implies, \[f(f(x_0)) = f((f(x_0)-x_0)+x_0) \ge f(f(x_0)-x_0)+x_0 > x_0\]which contradicts the second condition. Thus, for all $x \in \mathbb{R}^+$, $f(x) \le x$. Say there exists some positive real $x_0$ such that $f(x_0) < x_0$. Then, $P((x_0-f(x_0)),x_0)$ implies, \[f(x_0)=f((x_0-f(x_0))+f(x_0)) \ge f(x_0-f(x_0))+f(f(x_0))> f(x_0-f(x_0))\]which contradicts the strictly increasing nature of $f$. Thus, for all $x \in \mathbb{R}^+$, $f(x) \ge x$. Combining these two results, we have that the only possible solution is the desired one.