- #1
ArcanaNoir
- 779
- 4
Homework Statement
I'm actually only concerned here with proving equality. I would like some review of my proof before I crawl back to my professor again with what I think is a valid proof.
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.
Last edited: