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- #1

a^n - b^n <= n*(a^n-1)*(a-b) hypothesis

I used p(1) for a basis step. for the inductive step of p(k+1) I'm really at a loss here. Any thoughts on how I may tackle this, whatsoever, will be appreciated.

- Thread starter eric34
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- Thread starter
- #1

a^n - b^n <= n*(a^n-1)*(a-b) hypothesis

I used p(1) for a basis step. for the inductive step of p(k+1) I'm really at a loss here. Any thoughts on how I may tackle this, whatsoever, will be appreciated.

- Jan 30, 2012

- 2,577

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- #3

- Feb 7, 2012

- 2,807

Hi Eric34, and welcome to MHB.

a^n - b^n <= n*(a^n-1)*(a-b) hypothesis

I used p(1) for a basis step. for the inductive step of p(k+1) I'm really at a loss here. Any thoughts on how I may tackle this, whatsoever, will be appreciated.

For the inequality $a^n - b^n \leqslant na^{n-1}(a-b)$ to be true, I think you will have to assume that $a$ and $b$ are positive. The inequality can go wrong if $a$ or $b$ is allowed to be negative.

If you assume that $a\geqslant0$ and $b\geqslant0$, then the inequality is true. But I don't see how to prove it by using induction.

To prove it using algebra, you can factorise $a^n - b^n$ as $$a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2 + \ldots + b^{n-1}).$$ If $b\leqslant a$ then $a^{n-1}+a^{n-2}b+a^{n-3}b^2 + \ldots + b^{n-1} \leqslant na^{n-1}$, and so $a^n - b^n \leqslant na^{n-1}(a-b)$. On the other hand, if $a\leqslant b$ then $a^{n-1}+a^{n-2}b+a^{n-3}b^2 + \ldots + b^{n-1} \geqslant na^{n-1}$. Multiplying by $a-b$ (which is negative) changes the sign of the inequality so again we get $a^n - b^n \leqslant na^{n-1}(a-b)$.

To prove the inequality using calculus, divide both sides by $b^n$ and write $x=a/b$. The inequality then becomes $x^n-1\leqslant nx^{n-1}(x-1)$, or $(n-1)x^n - nx^{n-1}+1 \geqslant0.$ You can check that, for $x>0$, the function $f(x) = (n-1)x^n - nx^{n-1}+1$ has a minimum value of 0, when $x=1$, and is therefore positive for all other positive values of $x$.