# inequality proof

• Jan 30th 2010, 10:57 AM
Raoh
inequality proof
hello(Happy)
Let $(x_1,x_2,...,x_n)\in \mathbb{R}^{*}_{+}$,by applying this inequality, $\forall \alpha > 0,\log(\alpha )\leq \alpha -1$ to every value of,
$\alpha _i=\frac{x_i}{\frac{x_1+x_2+...+x_n}{n}},i=1,2,.., n.$
prove the following,
$\frac{1}{n}\sum_{k=1}^{n}\log x_k\leq \log(\frac{1}{n}\sum_{k=1}^{n}x_k)$
i believe it's the best place to post this question(Worried)
thanks.
• Jan 30th 2010, 12:14 PM
Black
For each $i$

$\text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right ) \le \frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}$.

As you combine the $n$ inequalities, the right side will cancel out, leaving you with

$\sum_{k=1}^{n}\log(x_i)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t) \le 0$.

Divide by $n$ to get the result.
• Jan 30th 2010, 12:29 PM
Raoh
• Jan 30th 2010, 01:31 PM
Black
We have the inequality $\text{log}(\alpha) \le \alpha - 1$ for nonnegative $\alpha$. So using $\alpha_i$ we have the left side of the inequality:

$\text{log}(\alpha_i)=\text{log}\left(\frac{x_i}{\f rac{x_1+\cdots+x_n}{n}}\right) = \text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right )$

and the right side:

$\alpha_i-1=\frac{x_i}{\frac{x_1+\cdots+x_n}{n}}-1=\frac{nx_i}{x_1+\cdots+x_n}-1=\frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}$.

There are $n$ of these inequalities since $i=1,2,\dots,n$. We add the $n$ inequalities together. For the left side, we will have

$\text{log}(x_1)+\cdots+\text{log}(x_n)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t)=\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t)$

and the right side:

$\frac{(n-1)x_1-\cdots-x_n}{x_1+\cdots+x_n}+\cdots+\frac{-x_1-\cdots+(n-1)x_n}{x_1+\cdots+x_n}=0$,

so

$
\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t) \le 0$

Divide both sides by $n$ to finish the proof.
• Jan 30th 2010, 01:35 PM
Raoh
Quote:

Originally Posted by Black
We have the inequality $\text{log}(\alpha) \le \alpha - 1$ for nonnegative $\alpha$. So using $\alpha_i$ we have the left side of the inequality:

$\text{log}(\alpha_i)=\text{log}\left(\frac{x_i}{\f rac{x_1+\cdots+x_n}{n}}\right) = \text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right )$

and the right side:

$\alpha_i-1=\frac{x_i}{\frac{x_1+\cdots+x_n}{n}}-1=\frac{nx_i}{x_1+\cdots+x_n}-1=\frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}$.

There are $n$ of these inequalities since $i=1,2,\dots,n$. We add the $n$ inequalities together. For the left side, we will have

$\text{log}(x_1)+\cdots+\text{log}(x_n)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t)=\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t)$

and the right side:

$\frac{(n-1)x_1-\cdots-x_n}{x_1+\cdots+x_n}+\cdots+\frac{-x_1-\cdots+(n-1)x_n}{x_1+\cdots+x_n}=0$,

so

$
\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ t) \le 0$

Divide both sides by $n$ to finish the proof.

thank you (Happy)