Whether $(a-k)^p-(a-h)^p\ge (h-k)^p$ for all $a>h>k>0 ,~ p\ge 1$?

Question

Whether $(a-k)^p-(a-h)^p\ge (h-k)^p$ for all $a>h>k>0 , ~p\ge1$ ?

I try some simple value, for example $a=3,h=2,k=1, p=2$ and so on. This inequality is right, but how to prove it ?

Answer 1

I think it's true.

$$(a-k)^p=(a-h+h-k)^p>(a-h)^p+(h-k)^p.$$ Indeed, let $a-h=x$ and $h-k=y$.

Thus, we need to prove that $$(x+y)^p>x^p+y^p.$$ Since the last inequality is symmetric of $x$ and $y$, we can assume that $x\geq y$.

Now, since $f(x)=x^p$ is a convex function and $(x+y,0)\succ(x,y)$, by Karamata we obtain: $$(x+y)^p+0^p>x^p+y^p$$ and we are done!