$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$ 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.

You can use the definition twice in a row; the trick is to choose values of $t$ that will produce $x+y$ and $x+z$ as arguments on the left-hand side. First take $t=\dfrac{z}{y+z}$, whence $1-t=\dfrac{y}{y+z}$:
$$
f\left(tx + (1-t)(x+y+z)\right) \le tf(x) + (1-t)f(x+y+z)\\
f\left(\frac{z}{y+z}x + \frac{y}{y+z}(x+y+z)\right) \le \frac{z}{y+z}f(x) + \frac{y}{y+z}f(x+y+z)
$$
The left-hand simplifies nicely because of the choice of $t$:
$$
f(x+y) \le \frac{z}{y+z}f(x) + \frac{y}{y+z}f(x+y+z)\tag1
$$
Now choose $t=\dfrac{y}{y+z}$ to get a similar inequality:
$$
f(x+z) \le \frac{y}{y+z}f(x) + \frac{z}{y+z}f(x+y+z)\tag2
$$
Adding $(1)$ and $(2)$ gives the desired result.
