$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$ for convex functions? For a convex function $f$, is it always true that
$$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$$ for all $x,y,z>0$?
I tried to use the definition of convexity: $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y).$$ but since the definition contains three occurrences of $f$ while the question contains four, I don't see how to plug in values to make them the same.

$f$ is convex then the graph of $f$ is below all its chords :
$$\forall z \in [x, y] \in \mathbb{R}, f(z) \leq f(x) + \dfrac{f(y) - f(x)}{y - x} (y - x)$$
Let $x, y, z > 0$.

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* We have $x < x + y < x + y + z$ then :
$$f(x + y) \leq f(x) + \dfrac{f(x + y + z) - f(x)}{(x + y + z) - x} ((x + y) - x) = f(x) + \dfrac{f(x + y + z) - f(x)}{y + z} y$$

* We have $x < x + z < x + y + z$ then :
$$f(x + z) \leq f(x + y + z) + \dfrac{f(x + y + z) - f(x)}{(x + y + z) - x} ((x + z) - (x + y + z)) = f(x + y + z) - \dfrac{f(x + y + z) - f(x)}{y + z} y$$

By summing the two inequalities :
$$f(x + y) + f(x + z) \leq f(x) + f(x + y + z)$$
