Prove inequality: $|x-y|^r \le 2^r (x^r+y^r)$

Question

This apparently holds by triangle inequality when $r>1$. But what about $0<r<1$? Any hint would be appreciated.

Answer 1

HINT: By the triangle inequality $|x-y|\leq|x|+|y|$.

Answer 2

It is equivalent to $$ \frac{|x - y|}{\sqrt[r]{|x|^r + |y|^r}} \leq 2.$$ By triangle inequality, the left hand side is bounded by $$ \frac{|x|}{\sqrt[r]{|x|^r + |y|^r}} + \frac{|y|}{\sqrt[r]{|x|^r + |y|^r}} \leq 1 + 1 = 2.$$