The easy(?) part of IMO 2011 Problem 3 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 $f(0) = 0$?
 A: 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.
A: First, we show that $f(x) \leq 0$ for all $x$.
Suppose that $f(z) > 0$ for some $z$. The functional inequality implies that $$\lim_{x \to -\infty} f(x) = -\infty,$$
since $f$ is bounded by a strictly increasing linear function.
We also have
$$ f(y) = f(0+y) \leq f(0)y + f(f(0)). $$
Using this for $y = f(x)$,
$$ f(0) = f(x-x) \leq -xf(x) + f(f(x)) \leq (f(0) - x)f(x) + f(f(0)).$$
As $x \to -\infty$, the right-hand side tends to $-\infty$, leading to a contradiction.
Second, let $x > 0$. Note that
$$ f(0) \leq x f(-x) + f(f(-x)) \leq xf(-x). $$
Thus $f(-x) \geq f(0)/x$. As $x\to\infty$, the righthand side tends to zero. Since
$$ f(-x) = f(-x+0) \leq f(f(-x)), $$
we get that there is a sequence of points $x_n = -f(-n)$ tending to zero such that $f(-x_n) \to 0$.
Suppose that $f(f(0)) < 0$. Then
$$f(-x_n) = f(0-x_n) \leq -f(0)x_n + f(f(0)).$$
Since $x_n\to 0$, the righthand side tends to $f(f(0)) < 0$, contradiction.
We conclude that $f(f(0)) = 0$. This implies that
$$0 = f(f(0)) \leq f(f(f(0))) = f(0), $$
and so $f(0) = 0$.
Third, as noted above, for $x > 0$ we have $$xf(-x) \geq f(0) = 0.$$
Thus $f(-x) = 0$ for all $x \geq 0$.
Addendum: Putting $g(x) = -f(x)$, the functional inequality reduces to
$$ g(x+y) \geq yg(x), $$
where now $g\colon \mathbb{R}_+ \to \mathbb{R}_+$. An example is the function $g(x) = \exp(x)$.
A: $f(0)=0$ is the last part of my solution :)
Step1: prove that $-xf(x)\ge 0$, for all $x$
Step2: prove that $f(x)\le 0$, for all $x$, then $f(x)=0$ for all $x<0$
Step3: $0=f(-1)=f(-1+0)\le 0$.
   $f(-1)+f(f(-1))=f(0)$, then $f(0)=0$.
