how to prove $\left | \sqrt[n]{a}- \sqrt[n]{b}\right |\leq \sqrt[n]{\left | a-b \right |}$

Question

$$\left | \sqrt[n]{a}- \sqrt[n]{b}\right |\leq \sqrt[n]{\left | a-b \right |},n\in \mathbb{N^{*}},a,b\geq 0\\$$

I failed to prove it, so if someone could prove it and explain it to me, it would be great.

Answer 1

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:

$$t^n\leq t^n + {n\choose 1}t^{n-1}+{n\choose 2}t^{n-2}+...t+1-1$$

Answer 2

Hint: maybe help you $${| \sqrt[n]{a}- \sqrt[n]{b}|=\\ | (\sqrt[n]{a}- \sqrt[n]{b})\times \frac{\sqrt[n]{a^{n-1}}+\sqrt[n]{a^{n-2}b^1}+\sqrt[n]{a^{n-3}b^2}+...+\sqrt[n]{b^{n-1}}}{\sqrt[n]{a^{n-1}}+\sqrt[n]{a^{n-2}b^1}+\sqrt[n]{a^{n-3}b^2}+...+\sqrt[n]{b^{n-1}}}|=\\ |\frac{a-b}{\sqrt[n]{a^{n-1}}+\sqrt[n]{a^{n-2}b^1}+\sqrt[n]{a^{n-3}b^2}+...+\sqrt[n]{b^{n-1}}}| } $$