Good evening, I was wondering how to prove easily this inequality: \begin{equation} py^{p-1}(x-y)\leq x^p-y^p\leq px^{p-1}(x-y) \end{equation} for all $x,y\geq0$ and $p\in[1,+\infty)$.
I tried to study both the functions $\frac{x^p-y^p}{y^{p-1}(x-y)}$ and $\frac{x^p-y^p}{x^{p-1}(x-y)}$, but I'm not getting anywhere.
For $x \geq y$ this follows immediately from MVT applied to the function $f(x)=x^{p}$ since $py^{p-1} \leq pt^{p-1} \leq px^{p-1}$ when $t \in [x,y]$. For $x<y$ just switch $x$ and $y$ and multiply the inequalities by $-1$.
A purely algebraic proof when $p$ is an integer, using high school algebra: we have the factorisation $$x^p-y^p=(x-y)(\underbrace{x^{p-1}+x^{p-2}y+\dots+xy^{p-2}+y^{p-1})}_{p\,\text{ terms}},$$ so that the second factor is $\le px^{p-1}\:$ and $\ge py^{p-1}\:$ if $\:0\le y\le x$. Multiplying by $x-y$ which is $\ge 0$ yields the result in this case.
If $0\le x\le y$, just permute the bounds for the second factor, and take into account that multiplying by $x-y\le 0$ yields again the inequalities.
