$(\lvert a\rvert +\lvert b\rvert)^{p}\leq 2^{p}(\lvert a\rvert^{p} +\lvert b\rvert^{p})$ for $p > 1$ and are absolute values necessary?

Question

Is it true that, for any $a,b\in \mathbb R$ and $p \geq 1$, we have

$(\lvert a\rvert +\lvert b\rvert)^{p}\leq 2^{p}(\lvert a\rvert^{p} +\lvert b\rvert^{p})$

If $p=2$ we have an explicit way to calculate by expansion,

$(\lvert a\rvert +\lvert b\rvert)^{2}=\lvert a\rvert^2+2\lvert a\rvert \cdot \lvert b\rvert +\lvert b\rvert^2 \leq 4\lvert a\rvert ^{2}+4\lvert b\rvert^{2}=2^2(\lvert a\rvert ^{2}+\lvert b\rvert^{2})$

How can I go about proving the others?

Is positivity necessary here?

Answer 1

$(|a|+|b|)^p\leq (2\max\{|a|,|b|\})^p=2^p(\max\{|a|,|b|\})^p\leq 2^p(|a|^p+|b|^p)$

The last step simply follows from the fact that $(\max\{|a|,|b|\})^p$ is either $|a|^p$ or $|b|^p$, so it is definitely not bigger than their sum. Here we indeed used that they are nonnegative.

Answer 2

More is true. The sharp inequality is

$$\tag 1(|a|+|b|)^p \le 2^{p-1}(|a|^p+|b|^p).$$

Proof: The function $x^p$ is convex on $[0,\infty)$ for $p\ge 1.$ Thus if $a,b\in \mathbb R,$ then $|a|,|b|\ge 0,$ and

$$\left (\frac{|a|+|b|}{2}\right )^p \le \frac{|a|^p}{2}+ \frac{|b|^p}{2}.$$

Multiplying by $2^p$ on both sides gives $(1).$