How do you prove the follwing $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$

Question

how do you prove this inequality? $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$ At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : $|x + ... -y| \le |x-y|$ I guess I would have to prove that ... $\le$ 0 but I don't know how to do that. Possibility 2 (n even) : $|x + ... +y| \le |x-y|$ Same as above but now I have to prove that ... $\le$ -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance.

Answer 1

Let $x\geq y$.

Thus, we need to probe that $$\left(\sqrt[n]x-\sqrt[n]y\right)^n\leq x-y$$ or $$\left(\sqrt[n]x-\sqrt[n]y+\sqrt[n]y\right)^n\geq\left(\sqrt[n]x-\sqrt[n]y\right)^n+\left(\sqrt[n]y\right)^n,$$ which is obvious for natural $n$.

Answer 2

hint

If $x\le y,$ it is true.

Assume $x>y=x\cos(t)$ with $0<t<\frac \pi2$.

After simplifying, we have to prove that

$$\Bigl(1-\cos^{\frac1n}(t)\Bigr)^n\le 1-\cos(t)$$

Answer 3

From the use of the $\sqrt[n]{\phantom{x}}$ sign, we are clearly* meant to assume that $x \geqslant 0$ and $y \geqslant 0$. If $x \geqslant y$, then $\sqrt[n]{x} \geqslant \sqrt[n]{y}$. If $x \leqslant y$, then $\sqrt[n]{x} \leqslant \sqrt[n]{y}$. By interchanging the roles of $x$ and $y$, if necessary, we can assume that $x \geqslant y$.

We now have to prove that $\sqrt[n]{x} - \sqrt[n]{y} \leqslant \sqrt[n]{x - y}$.

As both sides are non-negative, we needn't (yet!) worry whether $n$ is odd or even, we can raise everything to the power of $n$ as you suggested, and the inequality to be proved is equivalent to $\left(\sqrt[n]{x} - \sqrt[n]{y}\right)^n \leqslant x - y$.

It's easier to see what's going on if we change notation. Write $u = \sqrt[n]{x}$, $v = \sqrt[n]{y}$. We now have to prove that $(u - v)^n \leqslant u^n - v^n$ (where it is known that $u \geqslant v \geqslant 0$).

You might still think that now we have to worry about the parity of $n$. But we don't, and the reason is already shown in Michael Rozenberg's answer; so I'll stop here, just where his answer gets going.

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*I was a little hasty there. If $n$ is odd, we could have either $x < 0$ or $y < 0$, or both. But then the stated proposition is false, in general. For example, if $n = 3$, $x = 1$, and $y = -1$, it states that $|1 - (-1)| \leqslant \sqrt[3]{|1 - (-1)|}$, i.e. $2 \leqslant \sqrt[3]{2}$, which is false. So we must indeed assume that $x \geqslant 0$ and $y \geqslant 0$.