Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?

Question

Is

$$|x-y|^n\leq 2^n(|x|^n+|y|^n)$$

for all $x,y\in\mathbb{R}$ and $n=1,2,3,\dots$ a standard inequality? If so, what's its name or how do you prove it?

Answer 1

$|x|=(|x|^n)^{\frac{1}{n}}\le (|x|^n+|y|^n)^{\frac{1}{n}}$ and $|y|=(|y|^n)^{\frac{1}{n}}\le (|x|^n+|y|^n)^{\frac{1}{n}}$. So $$|x-y|\leq |x|+|y|\leq 2(|x|^n+|y|^n)^{\frac{1}{n}}$$ and consequently $|x-y|^n\leq 2^n(|x|^n+|y|^n)$

Answer 2

It is standard. I'm not sure if it has a name. Proof: $$|x-y|^n \le (|x| + |y|)^n \le (2 \max(|x|, |y|))^n = 2^n \max(|x|^n, |y|^n) \le 2^n (|x|^n + |y|^n).$$

Answer 3

Actually this inequality can be made stronger. It is true that

$$|x-y|^n\leq2^{n-1}(|x|^n+|y|^n),$$

with equality iff $x=-y$.

Note how your inequality follows from this because any real number is smaller then its double. This means your original inequality is weak only if $x=y=0$.

For a proof, we can start estimating the left side with the Triangle Inequality: $$|x-y|\leq|x|+|y|.$$

The inequality is then reduced to $$\frac{|x|+|y|}2\leq\sqrt[n]{\frac{|x|^n+|y|^n}2}$$

which is true because of the Power Mean Inequality. It can also be proven from AM-GM or the Rearrangement Inequality as follows:

$$\begin{align}2(|x|+|y|)^n&=\sum_{k=0}^n{n\choose k}\left(|x|^k|y|^{n-k}+|x|^{n-k}|y|^k\right)\\ &\leq\sum_{k=0}^n{n\choose k}\left(|x|^n+|y|^n\right)\\ &=2^n\left(|x|^n+|y|^n\right).\end{align}$$