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From Bulgaria '1998 there was a question,

Prove that there is no function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \begin{equation} \label{1} \tag{1} f^2(x) \geq f(x+y)\left(f(x) + y\right) \end{equation} 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 \begin{equation} \label{2} \tag{2} f(x) - f(x+y) \geq \frac{f(x)y}{f(x)+y} \end{equation}

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

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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.

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\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.

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