Estimate $|a+b|^p + |a-b|^p$

Question

I want to show, that for $a,b \in \mathbb{R}, p \in \mathbb{N}$ one has the following inequality: $$|a+b|^p + |a-b|^p \geq 2 |b|^p$$

I tried to use the binomial formula and triangle inequality, but got stuck on my way, as I only get $$\geq |(a+b)^p+(a-b)^p|= 2 \left|\sum_{k=0, k \text{ even}}^p \binom{p}{k} a^{p-k}b^k\right|$$.

How can I continue? Or am I on the wrong way?

Answer 1

Using the fact that the function $f(t)=|t|^p$ is convex, we have \begin{align} |b+a|^p+|b-a|^p &=2\big(\tfrac12f(b+a)+\tfrac12f(b-a)\big)\\ &\ge 2f\big(\tfrac12(b+a)+\tfrac12(b-a)\big)\\ &=2f(b) = 2|b|^p. \end{align} To see $f$ is convex, note $f''(t)=p(p-1)|t|^{p-2}\ge 0$. This suffices for all $p\ge 2$. For $p=1$, you can prove $f$ is convex using the triangle inequality, or just prove the inequality directly.

Answer 2

Hint:   using the generalized mean inequality, for $\,p \ge 1\,$:

$$ \sqrt[p]{\dfrac{|a+b|^p+|a-b|^p}{2}} \ge \dfrac{|a+b|+|a-b|}{2} $$

What's left to note is that $\,|a+b|+|a-b| \ge 2|b|\,$, which follows from the triangle inequality.