# How to prove that $|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1})$ when $x,y \geq 0$ and $p \geq 1$? [duplicate]

For $$x \geq 0$$, $$y \geq 0$$, prove that

$$|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).$$

I thought it would be simple but I messed everything up. Here are my attempts.

Fixing $$x > y \geq 0$$, I thought about the function

$$h(p)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})$$

Then $$h(1)=0$$ and I wanted to prove that $$h(p) \leq 0$$ when $$p \geq 1$$ by discussing its derivative but the derivative is messy offering no way out.

Also I thought about after fixing $$y \geq 0$$, consider the function

$$\varphi(x)=x^p-y^p-p(x-y)(x^{p-1}+y^{p-1})$$

where $$\varphi(y)=0$$ and I want to prove that $$\varphi(x) \leq 0$$ when $$x \geq y$$ yet again it's a messy way.

I also thought about restricting $$p$$ to $$\mathbb{N}$$ or $$\mathbb{Q}$$ using decomposition like

$$(x^p-y^p)=(x-y)(x^{p-1}+x^{p-1}y+x^{p-2}y^2+\cdots+y^{p-1})$$

but the number of terms does not match.

But I do believe this should be a easy question with some special background. I must have missed something critical. Any hint/solution appreciated!

• see if this helps with an induction proof $$x^p-y^p = (x-y)(x^{p-1}+y^{p-1}) + xy(x^{p-2}-y^{p-2})$$ – across Nov 28 '20 at 8:07
• @MartinR Yes, but in this page there are much more different solutions to check with! Anyway thank you for pointing out. – Zoe Desvl Nov 28 '20 at 11:18

Wlog $$x>y$$. Then with $$f(t):=x^p$$, we have from the Mean Value Theorem that there exists some $$\xi$$ with $$x>\xi>y$$ such that $$\frac{x^p-y^p}{x-y}=\frac{f(x)-f(y)}{x-y}=f'(\xi) =p\xi^{p-1}\le p\max\{x^{p-1},y^{p-1}\}

It's an direct consequence of the Hermite-Hadamard inequality

We have with $$f(x)=px^{p-1}$$ a convex function , $$x\geq y\geq 0$$ and $$p\geq 1$$:

$$\frac{1}{x-y}\int_{y}^{x}f(x)dx\leq \frac{f(x)+f(y)}{2}$$

Or :

$$\frac{x^p-y^p}{x-y}\leq \frac{px^{p-1}+py^{p-1}}{2}$$

WLOG $$x\ge y$$
let $$t=\frac{x}{y}\ge 1$$
we have to prove $$f(t)=(p-1)t^p-pt^{p-1}+pt-p+1\ge 0$$ notice $$f'(t)=p(p-1)(t^{p-1}-t^{p-2})+p>0$$ thus $$f(t)$$ is increasing also $$f(1)=0$$ hance $$f(t)\ge 0$$ for all $$x,y$$ in domain given (because the case $$x\le y$$ can be proved analogously)