Convex function $f(ax+by+cz) \leq af(x)+bf(y)+cf(z)$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a convex function. Let $x,y,z \in \mathbb{R}$ and $a,b,c \geq 0$ such that $a+b+c=1$. We want to show that $f(ax+by+cz) \leq af(x)+bf(y)+cf(z)$.

Obviously, if $c=0$, the result is equivalent to the definition of convexity. I think I should use the two variable case to prove the the three variable case (and ideally generalize to $n$ variables), but I don't see how to do it.

Answer 1

If $c > 0$ then $b + c > 0$ so $$ by + cz = (b + c)\left(\frac{b}{b + c}y + \frac{c}{b + c}z\right), $$ and $a + (b + c) = 1$. The motivation for this is that if you want to reduce to (usual) convexity then you should write $ax + by + cz = ax + (1-a)p$ where $p$ is some other point.

Answer 2

\begin{eqnarray*} f(ax+by+cz) &=& f \left( ax +(b+c) \frac{by+cz}{b+c} \right) \\ & \leq & a f(x) +(b+c) f \left( \frac{by+cz}{b+c} \right) \\ \end{eqnarray*} Now \begin{eqnarray*} f \left( \frac{by+cz}{b+c} \right) & \leq & \frac{b}{b+c} f ( y) + \frac{c}{b+c} f ( z) \\ \end{eqnarray*} Put these together and you have the result.