Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$py^{p-1}(x-y) \leq x^p-y^p \leq px^{p-1}(x-y)$$ 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.

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
This is the same as Didier's hint. – Yuval Filmus Jun 21 '11 at 18:06

Hint: If $y\le x$ and $m\le \varphi'(z)\le M$ for every $z\in[y,x]$, then $$(x-y)m\le \varphi(x)-\varphi(y)\le(x-y)M.$$

share|cite|improve this answer
Thanks for the hint! – user9352 Jun 21 '11 at 16:53

Hint: $$ \frac{x^p-y^p}{x-y} = x^{p-1} + x^{p-2}y + \cdots + x y^{p-2} + y^{p-1}. $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.