TITLE: The easy(?) part of IMO 2011 Problem 3
QUESTION [16 upvotes]: 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 $y$.
How can I prove that $f(0) = 0$?

REPLY [4 votes]: This is my solution:
$$f(f(x)) = f(y+f(x)-y) \le (f(x)-y).f(y) + f(f(y)),\forall x, y (1)$$
swap $x, y$, we have:
$$f(f(y))\le (f(y)-x)f(x)+f(f(x)), \forall x, y (2)$$
(1), (2) $\Rightarrow 0 \le 2f(x)f(y)-xf(x)-yf(y),  \forall x, y.$
$\Rightarrow -xf(x) \ge (y-2f(x))f(y), \forall x, y \Rightarrow -xf(x)\ge 0, \forall x (*)$ (using $y=2f(x)$)
In the other hand: $f(y)=f(x+y-x)\le (y-x)f(x)+f(f(x)), \forall x, y$. Suppose that there exist $x: f(x)>0$, then $\lim\limits_{y\to-\infty}(y-x)f(x)=-\infty \Rightarrow \lim\limits_{y\to-\infty}f(y)=-\infty \Rightarrow \lim\limits_{x\to-\infty}(-xf(x))=-\infty$! (absurd from $(*)$).
Therefore $f(x) \le 0, \forall x (**)$
From $(*), (**) \Rightarrow\forall x<0: -xf(x)\ge 0 \Rightarrow f(x)\ge 0 \Rightarrow f(x)=0.$
The last one: $0=f(-1)\le f(f(-1))=f(0)\Rightarrow f(0)=0.$
[End of proof]
koreagerman.