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I'm having trouble with the following problem:

Let u and v be real numbers such that $u > v > 0$ and prove by induction that for all $n \geq 2$, $u^{n} - v^{n} > (u - v)^{n}$.

I tried develop it as follows:

Base step (n=2):

$u^{2} - v^{2} > (u - v)^{2}$

$u^{2} - v^{2} > u^{2} -2uv + v^{2}$

$2uv > 2v^{2}$

(true since $u > v$)

Inductive step: Suppose that for some $k \geq 2$, $u^{k} - v^{k} > (u - v)^{k}$ is true, then follows that:

$u^{k+1} - v^{k+1} > (u - v)^{k+1}$

I stopped there. I really can't figure out how to conclude the inductive step. What should come next?

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  • $\begingroup$ You should have $v^2>uv$, not $v>uv$. $\endgroup$ – egreg Feb 19 '15 at 22:47
  • $\begingroup$ Yes, I have just fix it, thank you. $\endgroup$ – duarthiago Feb 19 '15 at 22:51
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$$(u-v)^{k+1}=(u-v)^k(u-v)<(u^k-v^k)(u-v)=u^{k+1}-u^k v-v^ku+v^{k+1}$$ but $-u^k<-v^k$, i.e. $$-u^kv<-v^kv=-v^{k+1}$$ and thus $$(u-v)^{k+1}<u^{k+1}-v^{k+1}-u^kv+\underbrace{v^{k+1}}_{=v^k v<u^kv}<u^{k+1}-v^{k+1}-u^kv+u^kv=u^{k+1}-v^{k+1}$$ what prove the claim.

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Set $w=u-v>0$, so you want to prove $$ (v+w)^n-v^n>w^n $$ or $$ (v+w)^n>v^n+w^n $$ The base case is obvious. Suppose it holds for $n$; then $$ (v+w)^{n+1}>(v^n+w^n)(v+w) $$ by the induction hypothesis. Can you go on?

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  • $\begingroup$ That just overkills the problem... XD $\endgroup$ – Patrick Da Silva Feb 19 '15 at 23:27
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    $\begingroup$ @PatrickDaSilva Why doing it the hard way when there's a comfortable path? ;-) $\endgroup$ – egreg Feb 19 '15 at 23:31
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$$u^{k+1}-v^{k+1}=u\left(u^k-\frac vu v^k\right)>u(u^k-v^k)>u(u-v)^k>(u-v)(u-v)^k$$

By the way. I don't think that your base step is right. $u>v$ does not imply that $v>uv$ (in fact, this latter inequality is false when $u\ge 1$). For this, I would write

$$u^2-2uv+v^2=u^2+v(v-2u)<u^2+v(v-2v)=u^2-v^2$$

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Write $u=v+(u-v)$. Suppose $u^n-v^n>(u-v)^n$ for $n\ge 2$, then $$u^{n+1}=u^n\cdot u>v^n\cdot u +(u-v)^n\cdot u>v^{n+1}+(u-v)^{n+1}$$ since $u>v>0$ and $u>u-v>0$.

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