For a natural number $n$ and numbers $a$ and $b$ such that $a \geq b \geq 0$, prove that $a^n-b^n \geq nb^{n-1}(a-b)$

Question

I tried to do an induction proof and I've played around with it for about an hour and haven't really gotten anywhere. For my base case, I let $n=1$ and got $(a-b) \geq (a-b)$, however when I tried to show $a^{k+1}-b^{k+1} \geq (k+1)b^{(k+1)-1}(a-b)$, I got stuck.

score 2 · Answer 1 · answered Sep 16 '14 at 16:57

2

Hint: Apply the mean value theorem on the function $f(x)=x^n$

answered Sep 16 '14 at 16:57

Marm

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A really clever way to prove this,well done! – Mathemagician1234 Sep 16 '14 at 18:27

score 1 · Answer 2 · answered Sep 16 '14 at 17:00

1

Hint:$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})$$

answered Sep 16 '14 at 17:00

Amitai Yuval

19,308

For a natural number $n$ and numbers $a$ and $b$ such that $a \geq b \geq 0$, prove that $a^n-b^n \geq nb^{n-1}(a-b)$

2 Answers2

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