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$$py^{p-1}(x-y) \leq x^p-y^p \leq px^{p-1}(x-y)$$ Where $0<y<x, \ 1\leq p <\infty$

I haven't been able to prove either of these inequalities. I tried subtracting the left from the middle and trying to show the whole thing is non-negative but haven't had any success.

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3 Answers

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It is a typical application of the Mean Value Theorem to a convex function (i.e. a function whose derivative f' is an increasing function).

Mean Value Theorem: if f is continuous and its derivative exists in the interval (a,b), then

f(b)-f(a) = f'(c)(b-a)

for some c in (a,b).

If f' is an increasing function, then a < c < b implies f'(a) < f'(c) < f'(b), but now (thanks to the Mean Value Theorem) you have an expression for f'(c). The last step is choosing the right convex function for your inequality. In fact, you can use the Mean Value Theorem to obtain many inequalities from many functions.

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This is the same as Didier's hint. –  Yuval Filmus Jun 21 '11 at 18:06
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Hint: If $y\le x$ and $m\le \varphi'(z)\le M$ for every $z\in[y,x]$, then $$(x-y)m\le \varphi(x)-\varphi(y)\le(x-y)M.$$

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Thanks for the hint! –  user9352 Jun 21 '11 at 16:53
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Hint: $$ \frac{x^p-y^p}{x-y} = x^{p-1} + x^{p-2}y + \cdots + x y^{p-2} + y^{p-1}. $$

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I see how the inequalities would follow from this but I don't see why this is true? Maybe I should point out that p could be an irrational # so maybe my original post title should be changed. I no longer think it is related to binomial expansion. See MVT post below. –  user9352 Jun 21 '11 at 16:51
    
To prove the hint, multiply the RHS by $x-y$ and see what you get. It might be possible to extend this argument to rational $p$, and then using a continuity argument you'll get it for arbitrary $p \geq 1$. –  Yuval Filmus Jun 21 '11 at 18:11
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