Does the inequality $|a-b|^p \geqslant 2^{-p}|a|^p - |b|^p$ hold for every $a,b \in \mathbb R$ and $1 \leqslant p < \infty$?

Question

I am wondering if the inequality

$$ |a-b|^p \geqslant 2^{-p}|a|^p - |b|^p $$

holds for every $a,b \in \mathbb R$ and every $1 \leqslant p < \infty$.

My attempt. First suppose that $|a| < |b|$. Then, since for every $p \in [1,\infty)$ we have that $2^{-p}|a|^p < |a|^p$, it follows that $2^{-p}|a|^p < |b|^p,$ which means that the right-hand side of the inequality is negative, and thus the inequality trivially holds.

Now let us deal with the case $|a| = |b|.$ In this case we have that

$$ 2^{-p}|a|^p - |b|^p = 2^{-p}|a|^p - |a|^p = \underbrace{|a|^p}_{\geqslant 0}\underbrace{(2^{-p}-1)}_{< 0} < 0, $$ which again means that the inequality trivially holds.

So all that's left is to deal with the case when $|a| > |b|$. I wasn't able to deal with this case bu I am highly confident that the inequality presented holds (I have elaborated a python program that test many different possibilities and all of them seem to be satisfied).

Can anyone help me with this last case?

Thanks for any help in advance.

Answer 1

$$ \begin{align*} \vert a \vert^p &\le (\vert a - b \vert + \vert b \vert)^p \\\\ &\le 2^p [\max(\vert a - b \vert, \vert b \vert)]^p \\\\ &\le 2^p \max(\vert a - b \vert^p, \vert b \vert^p) \\\\ &\le 2^p(\vert a - b \vert^p + \vert b \vert^p) \end{align*} $$ Here, note that $$ [\max(x, y)]^p = \max(x^p, y^p) \ \forall x, y > 0 $$ and $$ \max(x, y) \le x + y \ \forall x, y > 0 $$

Answer 2

The inequality in question is equivalent to

$$\lvert 2(a-b)\rvert^p + \lvert 2b\rvert^p\ge \lvert a\rvert^p,$$

which is true by Jensen's inequality since $x\mapsto \lvert x\rvert^p$ is convex for all $p\in[1,\infty)$.

In fact, even the stronger

$$\lvert 2(a-b)\rvert^p + \lvert 2b\rvert^p\ge 2\lvert a\rvert^p$$ is true.