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If $\sum_{j=1}^nx_j^2=1$ with $x_j\!\in\!\mathbb{R}$, why does it follow that $\sum_{j=1}^nx_j^4\geq\frac{1}{n}$. I'm trying to understand the following excerpt from Brandes & Erlebach's Network Analysis, p.407:

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up vote 3 down vote accepted

By generalized mean inequality (see e.g. Wikipedia, PlanetMath or AoPS) $$ \sqrt[4]{\frac{\sum_{i=1}^n x_i^4}n} \ge \sqrt[2]{\frac{\sum_{i=1}^n x_i^2}n}\\ \frac{\sum_{i=1}^n x_i^4}n \ge \left(\frac{\sum_{i=1}^n x_i^2}n\right)^2 $$ In your case $\sum x_i^2=1$ so you have $$\frac{\sum_{i=1}^n x_i^4}n \ge \frac1{n^2}\\ \sum_{i=1}^n x_i^4 \ge \frac1{n}.$$

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I was not aware of the generalized mean inequality, and the book didn't mention it anywhere. It's very simple when this inequality is known. Thank you! – Leon Aug 8 '12 at 5:47

By Cauchy-Schwarz inequality $$ 1=\sum_{j=1}^n x_j^2\cdot 1\leq \left( \sum_{j=1}^n (x_j^2)^2 \right)^{1/2} \left( \sum_{j=1}^n 1^2 \right)^{1/2}=\left( \sum_{j=1}^n x_j^4 \right)^{1/2} n^{1/2}. $$ Dividing by $n^{1/2}$ and squaring we obtain $$ \sum_{j=1}^n x_j^4\geq \frac{1}{n}. $$ We have equality if and only if $x_i=\pm\frac{1}{\sqrt{n}}$.

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Thanks for a different answer! CSB was all I really needed. – Leon Aug 8 '12 at 6:06

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