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I find the following problem interesting :

Find all functions $f \colon \Bbb R \rightarrow \Bbb R $ that satisfy the inequality
$f(x+y)+f(y+z)+f(z+x) \geq 3f(x+2y+3z)?$

How Can I tackle the problem? Any hints will be appreciated.

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    $\begingroup$ a source of the problem it is from 26th russian mathematical olympiad at 2000. Problem 9. equation-solver.blogspot.gr/2010/11/… $\endgroup$ Commented Apr 16, 2013 at 14:18
  • $\begingroup$ @JoséCarlosSantos with regards to your edit of my edit, I disagree that functional-inequalities should be independent of functional-equations. $\endgroup$ Commented Jun 29, 2019 at 10:43

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Plugging in $x=t,y=0,z=0$ yields $$f(0) \geq f(t), \tag{1}$$ and plugging in $x=t/2,y=t/2,z=-t/2$ yields $$f(t) \geq f(0). \tag{2}$$

Hence, from $(1)$ and $(2),$ we get $f(t)=f(0) \quad \forall t$,so $f$ must be constant.
Conversely, any constant function $f$ clearly satisfies the given condition.

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    $\begingroup$ (+1) I formatted it to look a little nicer. Check it out to see how I did it. $\endgroup$ Commented Apr 16, 2013 at 14:29
  • $\begingroup$ @CameronBuie thanks a lot for making it look nice. $\endgroup$ Commented Apr 16, 2013 at 14:35

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