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Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies
\[f(x + y) \leq yf(x) + f(f(x))\]
for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \leq 0$.

Easy for Imo 3. But since the solution is long, here are the main ideas only.

Firstly, set $y=0,f(x)\le f(f(x))$.
Then if $y=-x,f(0)\le f(f(x))-xf(x)$.
Set $y=f(x)-x,f(x)(f(x)-x)\ge0(*)$.
Now if $y=f(0)-x,f(f(0))\le f(f(x))+f(0)f(x)-xf(x)\to(1)$
Again set $x=0,y=f(x)$ and use $(1)$ to show $2f(0)f(x)\ge xf(x)$. Consider $f(x)\ge0,x\le2f(0)$, use $(*)$ and if $f(x)\le0,x\ge2f(0)\forall x$, contradiction. So $f(x)=0$.

Let $P(x,y)$ be the given assertion.
Comparing $P(x,f(y)-x)$ and $P(y,f(x)-y)$ yields, $$xf(x)+yf(y)\leq 2f(x)f(y).$$
$y\mapsto 2f(x)\Rightarrow xf(x)\leq 0. \qquad (*)$
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$\textbf{Claim: }f(k)\leq 0~~\forall k.$
$Proof.$Suppose $\exists k:f(k)>0,$ then $$f(k+y)\leq yf(k)+f(f(k)).$$
Now $y\to -\infty$ implies that $\lim_{x\to -\infty} f(x)=-\infty.$
$P(x,z-x)\Rightarrow f(z)\leq (z-x)f(x)+f(f(x)).$
Then $x\to -\infty,$ yields a contradiction. $\blacksquare$
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From $(*)$ we get $f(x)=0,\forall x<0.$
$P(0,f(0))\Rightarrow f(0)\geq 0,$ thus we get $f(0)=0,$ as desired.