(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'm actually only concerned here with provingequality. I would like some review of my proof before I crawl back to my professoragainwith what I think is a valid proof.

3. The attempt at a solution

Show:

[tex] \frac{x_1+x_2+...+x_n}{n}=\sqrt[n]{x_1x_2\cdots x_n} \Leftrightarrow x_1=x_2=\dots=x_n [/tex]

Given:

[tex] \frac{x_1+x_2}{2}=\sqrt{x_1x_2} \Leftrightarrow x_1=x_2 [/tex]

Assume it is true for [itex] n=k [/itex]

That is, assume:

[tex] \frac{x_1+x_2+...+x_k}{k}=\sqrt[k]{x_1x_2\cdots x_k} \Leftrightarrow x_1=x_2=\dots=x_k [/tex]

for [itex] n=k+1 [/itex] we have:

[tex] \frac{x_1+x_2+...+x_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_1x_2\cdots x_kx_{k+1}} [/tex]

Because of the assumption for [itex]k[/itex], we can write:

[tex] \frac{kx_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_k^kx_{k+1}} [/tex]

let [itex] x_k-x_{k+1}=\delta [/itex]

now we can replace [itex]x_k[/itex] by [itex] (x_{k+1}+\delta) [/itex] on one side and [itex] x_{k+1} [/itex] by [itex] (x_k-\delta) [/itex] on the other:

[tex] \frac{kx_k+x_k-\delta}{k+1}=\sqrt[k+1]{(x_{k+1}+\delta)^kx_{k+1}} [/tex]

[tex] \lim_{\delta \rightarrow 0} \frac{kx_k+x_k-\delta}{k+1}=\lim_{\delta \rightarrow 0} \sqrt[k+1]{(x_{k+1}+\delta)^kx_{k+1}} [/tex]

[tex]

\frac{(k+1)x_k}{k+1}=\sqrt[k+1]{(x_{k+1})^kx_{k+1}}

[/tex]

[tex] x_k=x_{k+1} [/tex]

Thus, equality holds iff [itex] x_1=x_2=\dots =x_n=x_{n+1} [/itex]

By the Principle of Mathematical Induction, the proof is over.

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# Homework Help: Please review my proof of Cauchy inequality

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