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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!

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

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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 :

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

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  • $\begingroup$ very useful,i never knew about this inequality(+1) $\endgroup$ – Albus Dumbledore Nov 28 '20 at 10:15
  • $\begingroup$ btw does the reverse hold if it is concave $\endgroup$ – Albus Dumbledore Nov 28 '20 at 10:17
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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)

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