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An inequality

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Riemann
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An inequality

#1

Post by Riemann » Thu Nov 22, 2018 9:26 pm

Let $x_1, x_2, \dots, x_n$ be $n \geq 2$ positive numbers other than $1$ such that $x_1^2+x_2^2+\cdots +x_n^2=n^3$. Prove that:

$$\frac{\log_{x_1}^4 x_2}{x_1+x_2}+ \frac{\log_{x_2}^4 x_3}{x_2+x_3}+ \cdots + \frac{\log_{x_n}^4 x_1}{x_n+x_1} \geq \frac{1}{2}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Riemann
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Re: An inequality

#2

Post by Riemann » Wed Oct 23, 2019 7:58 pm

The Engels form of the Cauchy – Schwartz inequality gives us:

\begin{align*}
\sum \frac{\log_{x_1}^4 x_2}{x_1+x_2} & \geq \frac{\left (\sum \log_{x_1}^2 x_2 \right )^2}{\sum (x_1+x_2)} \\
&= \frac{\left ( \sum \log_{x_1}^2 x_2 \right )^2}{2\sum x_1} \\
&\!\!\!\!\!\!\overset{\text{AM-GM}}{\geq } \frac{\left [ n \left (\prod \log_{x_1} x_2 \right )^{2/n} \right ]^2}{2\sum x_1} \\
&\!\!\!\!\overset{\text{C-B-S}}{\geq } \frac{n^2}{2n^2} \\
&= \frac{1}{2}
\end{align*}

and the inequality is proved!
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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