Prove that for $1\leq p\leq 2$ and $0<b<a$, $(a+b)^p+(a-b)^p\geq 2a^p+p(p-1)a^{p-2}b^2$.

Question

To start this one, my idea is dividing both side by $a^p$, then the question will become $(1+x)^p+(1-x)^p\geq 2+p(p-1)x^2$. So we need to check $f(x)=(1+x)^p+(1-x)^p- 2-p(p-1)x^2\geq 0$ on $x\in(0,1)$ for all $p\in[1,2]$. It's clear that it's true as $p=1,2$. And for $p\in(0,1)$, I try to take first and second derivative, but the case becomes very complicate.

Answer 1

No, it's not complicated: $$f''(x)=p(p-1)((1+x)^{p-2}+(1-x)^{p-2}-2)\geq0$$ by Jensen.

Can you end it now?

I used the following Jensen: $$\frac{(1+x)^{p-2}+(1-x)^{p-2}}{2}\geq\left(\frac{1+x+1-x}{2}\right)^{p-2}=1.$$