TITLE: how to prove $\left | \sqrt[n]{a}- \sqrt[n]{b}\right |\leq \sqrt[n]{\left | a-b \right |}$
QUESTION [5 upvotes]: $$\left | \sqrt[n]{a}- \sqrt[n]{b}\right |\leq \sqrt[n]{\left | a-b  \right |},n\in \mathbb{N^{*}},a,b\geq 0\\$$
I failed to prove it,  so if someone could prove it and explain it to me, it would be great.

REPLY [7 votes]: Say $x=\sqrt[n]{a}$ and $y=\sqrt[n]{b}$. Then we have to prove:
$$ |x-y|^n \leq |x^n-y^n|$$ Because of simetry we can assume that $x\geq y$. Let $z={x\over y}$. Since $z\ge 1$ we have to prove:
$$ (z-1)^n \leq z^n-1$$
Now write $t=z-1$, so we have to prove $$t^n\leq (t+1)^n-1 $$ which is true by binomial theorem:
$$t^n\leq t^n + {n\choose 1}t^{n-1}+{n\choose 2}t^{n-2}+...t+1-1$$