Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$

I got stuck on this inequality for a day. If $p$ is positive integer, then the problem becomes too easy, but I can't find how we deal with the general case when p can be any positive number. Can someone give me a hint? I really appreciate

Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$

WLG, assume $$x, by theorem we have, $$|x^p - y^p|/|x-y| = \frac{d}{du}(u^p)(c)=pc^{p-1} , c \in [x,y].$$
$$c^{p-1} \le (x^{p-1} + y^{p-1}).$$
Multiply both sides by $|x-y| and the theorem follows. • Simple and beautiful. Thanks so much. This inequality is a small part from Real and Complex Analysis of Walter Rudin, page 75 in$L^p$space chapter. So I don't think it could be wrong! – le duc quang Jul 27 '16 at 15:29 • I think it could be simple like this: For every$p$, the function$f(x) = x^p$is either increasing or decreasing, so because$c \in [x,y]$and$x, y \ge 0$, we always have$c^{p-1} \le (x^{p-1} + y^{p-1})\$ – le duc quang Jul 27 '16 at 15:37