Prove an inequality with absolute values I need to prove the following inequality, because I have to use it for a problem I'm solving. 
For $x,y \geq 0$
$ 0 < p < 1,$ then $|x^p-y^p| \leq |x-y|^p$
$ 1 \leq p < \infty,$ then $|x^p-y^p| \leq p |x-y|(x^{p-1} + y ^{p-1})$ 
I have tried using the Newton's formula expanding $(x-y)^p$, but I didn't get anywhere. Any ideas how I could do it?
 A: By symmetry, for both inequalities, if it's true for $x,y$, then it's also true for $y,x$. Thus, WLOG, let $x \ge y$. Consider the first inequality you wish to prove, i.e.,
$0 < p < 1,$ then $|x^p-y^p| \leq |x-y|^p$
This is obviously true for where $x = y$. For any given $y$, and for all $x \ge y$, since $f(z) = z^p$ is an increasing function for $z \ge 0$, you can remove the absolute value signs and move the right side value to the left to get a new function you wish to prove of
$$f(x) = (x - y)^p - (x^p - y^p) \ge 0 \tag{1}\label{eq1A}$$
Taking the derivative gives
$$\begin{equation}\begin{aligned}
f'(x) & = p(x - y)^{p-1} - px^{p-1} \\
& = p\left((x - y)^{p-1} - x^{p-1}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $p \lt 1$, then $p - 1 \lt 0$. Also, considering $x \gt y$, you have $0 \lt x - y \lt x$, so $(x - y)^{1-p} \lt x^{1-p} \implies (x - y)^{p-1} \gt x^{p-1}$. Thus, the derivative in \eqref{eq2A} is positive for $x \gt y$, so since $f(y) = 0$, this means \eqref{eq1A} is true for all $x \ge y$, proving your original inequality.
You can do something similar with your second inequality, which I'll leave to you to finish yourself.
