# Inequality of weights on a graph

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:

-

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 $$\frac{\sum_{i=1}^n x_i^4}n \ge \frac1{n^2}\\ \sum_{i=1}^n x_i^4 \ge \frac1{n}.$$
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 $x_i=\pm\frac{1}{\sqrt{n}}$.