# Prove that there is no function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f^2(x) \geq f(x+y)(f(x) + y)$

From Bulgaria '1998 there was a question,

Prove that there is no function $$f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$$ such that $$$$\label{1} \tag{1} f^2(x) \geq f(x+y)\left(f(x) + y\right)$$$$ for all $$x,y \in \mathbb{R}^+$$.

In Titu Andreescu's book the solution is provided as linked (half-solution) here, where he makes a manipulation to frame the inequality $$$$\label{2} \tag{2} f(x) - f(x+y) \geq \frac{f(x)y}{f(x)+y}$$$$

I fail to see how (2) is derived from the given inequality (1) or verify how so.

The given inequality is $$f(x+y) \leq \frac {f^{2}(x)} {f(x)+y}$$. So $$f(x)-f(x+y) \geq f(x)- \frac {f^{2}(x)} {f(x)+y}$$. Now simplify the right hand side.
\begin{align}[f(x)-f(x+y)](f(x)+y)&=f^2(x)+f(x)y-f(x+y)[f(x)+y]\\ &=f(x)y + [f^2(x)-f(x+y)(f(x)+y)]\\ &\ge f(x)y \end{align}
where the last inequality is due to $$(1)$$.
Dividing both sides by $$f(x)+y$$ which is positive gives you the result.