Prove that $(|u-s|+|x-y|)^2\leq 2|u-s|^2+2|x-y|^2$.

Question

Prove that $(|u-s|+|x-y|)^2\leq 2|x-y|^2+2|u-s|^2$.

My professor used this inequality for a proof last week. How would one prove this? I thought about using the Cauchy-Swartz inequality.

This is not a homework question. I am interested in the proof just for self-learning.

Answer 1

$$(a+b)^2\le (a+b)^2+(a-b)^2 = (a^2+2ab+b^2)+(a^2-2ab+b^2)= 2a^2+2b^2.$$