$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$ 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.
 A: 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.
A: I agree with Tintarn about this inequality being a straightforward consequence of Karamata's inequality, but it is also quite easy to prove from scratch. 
Given a convex and differentiable function $f$, if we define $\Delta_f$ as:
$$ \Delta_f(a,b)=\left\{\begin{array}{rcl}\frac{f(a)-f(b)}{a-b}&\text{if}&a\neq b,\\ f'(a)&\text{if}&a=b\end{array}\right.\tag{1}$$
we have that $\Delta_f$ is a weakly increasing function with respect to both its parameters.
Our inequality is equivalent to:
$$\Delta_f(x+y+z,x+y)\geq \Delta_f(x+z,x)\tag{2} $$
and since $x,y,z>0$ we have that $(2)$ just follows from $(1)$ (differentiability does not really matter, since we do not really need to define $\Delta_f(a,a)$ in order to prove $(2)$ through $(1)$).
A: $f$ 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 $x, y, z > 0$.

*

* 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 :
$$f(x + y) + f(x + z) \leq f(x) + f(x + y + z)$$
