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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 $x, y \ge 0$

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WLG, assume $x<y$, by theorem we have, $$|x^p - y^p|/|x-y| = \frac{d}{du}(u^p)(c)=pc^{p-1} , c \in [x,y].$$

But thanks to the comment by @le duc quang below,

$$c^{p-1} \le (x^{p-1} + y^{p-1}).$$

Multiply both sides by $|x-y| and the theorem follows.

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  • $\begingroup$ 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! $\endgroup$ – le duc quang Jul 27 '16 at 15:29
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    $\begingroup$ 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 $c^{p-1} \le (x^{p-1} + y^{p-1})$ $\endgroup$ – le duc quang Jul 27 '16 at 15:37

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