Find all functions $f:Q\rightarrow Q$ such that \[ f(x+y)+f(y+z)+f(z+t)+f(t+x)+f(x+z)+f(y+t)\ge 6f(x-3y+5z+7t) \] for all $x,y,z,t\in Q.$
Problem
Source: Uzbekistan NO 2013, P3
Tags: function, algebra proposed, algebra
27.05.2013 08:05
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.
27.05.2013 18:17
I thin it is the esiest problem in UZB nmo 2013!
27.05.2013 18:20
Hey mathuz Are you from uzbekistan!?
27.05.2013 18:26
The condition $ f:Q\rightarrow Q $ is realy useless
27.05.2013 19:06
No; it wouldn't have worked if $f:\mathbb{Z}\to\mathbb{Z}$
28.05.2013 00:08
But if the function is from R to R it is basically the same
28.05.2013 12:42
$f(x)=const$
25.12.2013 13:41
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 .