What a fun and easy functional equation for a 3rd problem Let $P(a,b)$ be the given equation. $P(a,b)$ for a large enough $a$ $\implies$ $f(x)$ has an infinitely small value. Let $f(c)$ be small enough. Then $P(a,{c}/{f(a)})$ means that $f$ is surjective. Let $f(bf(a))+a=k$ for some parameters $a$ and $k$, and variable $b$. This equation has solution $\iff$ $a<k$, since $f$ is surjective. This means that $Q(a,k) = f(af(b)+a) = 1/k$ also has a solution (for the same values of $a$ and $k$) $\iff$ $a<k$. We also have that $af(b)+a = a(f(b)+1)$, takes on all values larger than $a$, again due to surjectivity. Let $f(z) = 1/k$. $Q(k,k)$ doesn't have a solution $\implies z \leq k$. But $Q(k-\epsilon,k)$ does have a solution $\implies z > k-\epsilon \implies z \geq k \implies z=k \implies \boxed{f(k) = 1/k}$ which clearly works.

Remark

Let $P(a,b)$ the assertion of the F.E. Claim 1: $\lim_{x \to \infty} f(x)=0$ Proof: By $P \left(\frac{a}{f(b)+1}, b \right)$ we have: $$f(a)=\dfrac{1}{f \left(b f \left(\frac{a}{f(b)+1} \right) \right)+\frac{a}{f(b)+1}}<\frac{f(b)+1}{a}$$This holds for any $a,b \in \mathbb R^+$ so fix $b$ and take $a$ really huge and by definition of the limit at infinity our claim is proven. Claim 2: $xf(x) \le 1$ for all $x \in \mathbb R^+$ Proof: In the inequality from Claim 1 just set $b \to \infty$ and done. Claim 3: $f$ is surjective. Proof: By $P \left(a-f(b), \frac{b}{f(a)} \right)$ for $a>f(b)$ we have: $$f \left((a-f(b)) \left(f \left(\frac{b}{f(a)} \right)+1 \right) \right)=\frac{1}{a}$$As we can make $f(b)$ arbitrarily small we have proven this claim. Claim 4: $f$ is injective. Proof: Suppose $f(m)=f(n)$ for $m>n$ then by $P(a,m)-P(a,n)$ and Claim 3 we can easly get $f(mx)=f(nx)$ for all $x \in \mathbb R^+$, now this means: $$f(x)=f \left(\left(\frac{m}{n} \right)^N \cdot x \right) \le \dfrac{1}{\left(\frac{m}{n} \right)^N \cdot x} \overset{N \to \infty}{\implies} f(x) \le 0 \; \text{contradiction!}$$Finishing: Clearly for $c>a$ there exists a $b$ such that for such $a$ and fixed $c$ we have $f(af(b)+a)=\frac{1}{c}=f(z)$ (because by $P(a,b)$ we get $f(bf(a))=c-a$ so we choose a suitable $b$ for any $a<c$) , note also $c \ge z>a$ (because $z=af(b)+a>a$ and $\frac{1}{c} \le \frac{1}{z}$) then , but a suitable $b$ lets us choose $a=c-\epsilon$ for any $\epsilon$ really really small so $c \ge z>c-\epsilon$ which in fact would mean by trivial analysis stuff that $c \ge z \ge c$ therefore $c=z$ and $f(c)=\frac{1}{c}$, since in this process $c$ was not bounded or anything... The only solution to this F.E. is $\boxed{f(x)=\frac{1}{x} \; \forall x \in \mathbb R^+}$ thus we are done

Easy functional equation by surjectivity

The answer is $\boxed{f\equiv 1/x}$, which can be verified to work: Claim: $\lim_{x\rightarrow \infty}f(x)=0$ Taking $a\rightarrow \infty$ this follows by the sandwich theorem. Claim: $f$ is surjective Notice that $1/(f(bf(a))+a)$ is in the domain of $f$, by varying $a$ and choosing $bf(a)$ to be a fixed large number finishes. Claim: $f(x)\leq 1/x$ Taking $a=x/(1+f(b))$ have that $f(x)\leq (1+f(b))/x$ so simply take $b\rightarrow \infty$. Claim: $f$ is injective If $f(b)=f(c)$ with $c<b$ then $f(bf(a))=f(cf(a))$ or $f(\frac{b}{c}x)=f(x)$ or that $f(\frac{b^n}{c^n}x)=f(x)$, contradicting the first claim. Fix $a$. Now notice that $f(af(b)+a)$ achieves all values smaller than $\frac{1}{a}$ or that $f:(a,\infty)\rightarrow (0,\frac{1}{a})$ is a bijection. However $f:(a+\epsilon,\infty)\rightarrow (0,\frac{1}{a+\epsilon})$ is also a bijection so $f:(a,a+\epsilon)\rightarrow(\frac{1}{a+\epsilon},\frac{1}{a})$ is also a bijection. Taking $\epsilon\rightarrow 0$ and varying $a$ finishes the problem.

Let $P(x,y)$ be the assertion $f\left( xf\left( y \right)+x \right)=\frac{1}{f\left( yf\left( x \right) \right)+x}$ 1) $\lim_{x\rightarrow +\infty}f(x)=0$

Proof

2) $f\left( x \right)\le \frac{1}{x},\forall x>0$

Proof

3)$f$ is surjective

Proof

4) $f$ is injective

Proof

5) $f\left( a \right)<1\Rightarrow a>1$

Proof

5) $y>z\Rightarrow f(y)<f(z)$

Proof

6) $\boxed{f\left( x \right)=\frac{1}{x},\forall x>0}$

Proof

@topologicalsort wrote Quote: Let $P(a,b)$ be the given equation. $P(a,b)$ for a large enough $a$ $\implies$ $f(x)$ has an infinitely small value. Let $f(c)$ be small enough. Then $P(a,{c}/{f(a)})$ means that $f$ is surjective. Let $f(bf(a))+a=k$ for some parameters $a$ and $k$, and variable $b$. This equation has solution $\iff$ $a<k$, since $f$ is surjective. This means that $Q(a,k) = f(af(b)+a) = 1/k$ also has a solution (for the same values of $a$ and $k$) $\iff$ $a<k$. We also have that $af(b)+a = a(f(b)+1)$, takes on all values larger than $a$, again due to surjectivity. Let $f(z) = 1/k$. $Q(k,k)$ doesn't have a solution $\implies z \leq k$. But $Q(k-\epsilon,k)$ does have a solution $\implies z > k-\epsilon \implies z \geq k \implies z=k \implies \boxed{f(k) = 1/k}$ which clearly works. Except for the last two lines of the solution, it shows that $f^{-1}(1/k) \cap (a,\infty) \neq \emptyset \iff a<k$. However, we also need to exclude the possibility that $f^{-1}(1/k)$ has $k$ as a limit point without $k$ being a member.