Let $P(x,y,z,t)$ be the assertion. $P(x,0,0,0)\implies f(0)\ge f(x)$. $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2},\frac{x}{2})\implies f(x)\ge f(0)\implies f(x)=f(0)\;\forall x\in\mathbb{Q}$ which is indeed a solution.

I thin it is the esiest problem in UZB nmo 2013!

Hey mathuz Are you from uzbekistan!?

The condition $ f:Q\rightarrow Q $ is realy useless

No; it wouldn't have worked if $f:\mathbb{Z}\to\mathbb{Z}$

But if the function is from R to R it is basically the same

$f(x)=const$

Faustus wrote: Let $P(x,y,z,t)$ be the assertion. $P(x,0,0,0)\implies f(0)\ge f(x)$. $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2},\frac{x}{2})\implies f(x)\ge f(0)\implies f(x)=f(0)\;\forall x\in\mathbb{Q}$ which is indeed a solution. I think this is good solution .