Assume contrary that $f(x)<N$ for all $x\in\mathbb{R}^+$. $$f(x+f(y))^2\geq f(x)\left(f(x+f(y))+f(y)\right)>f(x)\cdot f(x+f(y))$$$\implies f(x+f(y))>f(x)$. So $$N^2>f(x+f(y))^2\geq f(x)\left(f(x+f(y))+f(y)\right)>f(x)^2+f(x)\cdot f(y)$$Setting $x=y$ gives that $\frac{N}{\sqrt{2}}>f(x)$ for all $x\in\mathbb{R}^+$. So induction give that $\frac{N}{(\sqrt{2})^n}>f(x)$ for all $n\in\mathbb{N}$. Sending $n$ to infinity gives contradiction. So $f$ is unbounded.

if $\mathbb{R}^+$ refers to the set of nonnegatives reals as french meaning we have to add $ f\ne 0$

Put $x=y$ we get $f(x+f(x))^2 \ge f(x) \left( f(x+f(x))+f(x) \right)$, so $f(x+f(x)) \ge \frac{1+\sqrt{5}}{2} f(x)$. Iterating shows that $f$ is unbounded.

$P(x,x) \implies f(x+f(x)) > f(x)$ it follows from this inequality that the function $f$ is unbounded.

lazizbek42 wrote: $P(x,x) \implies f(x+f(x)) > f(x)$ it follows from this inequality that the function $f$ is unbounded. I don't think so. Because $f(m) >f(n) $ for all $m>n$ doesn't mean that this function is unbounded. For example $f(x) =1-\frac{1}{2^x}$ is strictly increasing function, but it has bound, which is $1$.