Prove the following inequality $\sum_{k=0}^{n}(-1)^{k}f(a_{k})\geq f ( \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } a _ { k } )$ Suppose that a function $f$ is convex and increasing on $[0,+\infty)$ and $f(0)=0$ .Show that
$$\sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } f ( a_ { k } ) \geq f \left( \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } a _ { k } \right)$$
For any number $a _ { 0 } \geq a _ { 1 } \geq \ldots \geq a _ { n } \geq 0$

Please help me to solve this I have try using Jensen’s inequality.
But I have thought it many times. I still can’t do this.
Thank beforehand!

It is enough to prove for even numbers $n=2m$, since for an odd number of $a_0,...,a_{2m-1}$ we can make it even by adding $a_{2m}=0$ to the set of numbers.
The induction start ($m=1$) with $a_0\geq a_1 \geq a_2$ follows from
$$\int_{a_1-a_2}^{a_0} f'(t-a_1+a_2) \, {\rm d}t = f(a_0-a_1+a_2) - f(0) \\
=\int_{a_1-a_2}^{a_1} f'(t-a_1+a_2) \, {\rm d}t + \int_{a_1}^{a_0} f'(t-a_1+a_2) \, {\rm d}t \\
\leq f(a_2) - f(0) + \int_{a_1}^{a_0} f'(t) \, {\rm d}t \\
=f(a_0) - f(a_1) + f(a_2)$$
since $f$ is convex with $f(0)=0$.
It then follows after going from $m$ to $m+1$ $$\sum_{k=0}^{2m+2} (-1)^k f(a_k) = f(a_0) - f(a_1) + \sum_{k=2}^{2m+2} (-1)^k f(a_k) \\
\geq f(a_0) - f(a_1) + f\left(\sum_{k=2}^{2m+2} (-1)^k a_k\right)\\
\geq f\left(a_0 - a_1 + \sum_{k=2}^{2m+2} (-1)^k a_k\right) = f\left( \sum_{k=0}^{2m+2} (-1)^k a_k \right) \, $$
since $(-1)^2=1$ (allowing the shift and applying the induction hypothesis) and $\sum_{k=2}^{2m+2} (-1)^k a_k \leq a_2 \leq a_1 \leq a_0$.
