- #1
ArcanaNoir
- 778
- 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:
\frac{x_1+x_2+...+x_n}{n}=\sqrt[n]{x_1x_2\cdots x_n} \Leftrightarrow x_1=x_2=\dots=x_n
Given:
\frac{x_1+x_2}{2}=\sqrt{x_1x_2} \Leftrightarrow x_1=x_2
Assume it is true for n=k
That is, assume:
\frac{x_1+x_2+...+x_k}{k}=\sqrt[k]{x_1x_2\cdots x_k} \Leftrightarrow x_1=x_2=\dots=x_k
for n=k+1 we have:
\frac{x_1+x_2+...+x_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_1x_2\cdots x_kx_{k+1}}
Because of the assumption for k, we can write:
\frac{kx_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_k^kx_{k+1}}
let x_k-x_{k+1}=\delta
now we can replace x_k by (x_{k+1}+\delta) on one side and x_{k+1} by (x_k-\delta) on the other:
\frac{kx_k+x_k-\delta}{k+1}=\sqrt[k+1]{(x_{k+1}+\delta)^kx_{k+1}}
\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}}
<br /> \frac{(k+1)x_k}{k+1}=\sqrt[k+1]{(x_{k+1})^kx_{k+1}}<br />
x_k=x_{k+1}
Thus, equality holds iff x_1=x_2=\dots =x_n=x_{n+1}
By the Principle of Mathematical Induction, the proof is over.
Last edited:
