Solve $f (x + y) + f (y + z) + f (z + x) \ge 3f (x + 2y + 3z)$

Find all functions $f : \mathbb{R} \to \mathbb{R}$ which satisfy :

$f (x + y) + f (y + z) + f (z + x) ≥ 3f (x + 2y + 3z)$

for real $x,y,z$.

Attempt at solution:

I have tried plugging in $x = -y$ and $x = -z$. This does not seem to be getting me anywhere.

Any help is appreciated.

Solution 1:

Any constant function would meet the condition.

If we plug in $x = a$, $y = 0$, $z = 0$, we get $f (a) + f (0) + f (a) \geqslant 3f (a)$, which implies that $f(0)\geqslant f (a)$.

Plugging in $x = a/2$, $y = a/2$, $z = −a/2$ gives us $f (a) + f (0) + f (0) \geqslant 3f (0)$, which implies that $f(a)\geqslant f (0)$.

Hence $f (a) = f (0)$ for all real $a$, so $f$ must be constant.