# prove that $\frac{\sqrt{m}}{m} \sum_{i=1}^m |x_i| \leq |x| \leq \sum_{i=1}^m |x_i|$

Let $$x=(x_1,...x_m) \in R^m$$, prove that $$\frac{\sqrt{m}}{m} \sum_{i=1}^m |x_i| \leq |x| \leq \sum_{i=1}^m |x_i|$$, where $$|x|=\sum_{i=1}^m x_i$$

The proof goes by the following:

Since $$(\sum_{i=1}^m |x_i|)^2=\sum_{i=1}^m x_i^2+2 \cdot \sum_{i=1_{m\geq j >i}}^m |x_i| |x_j| \leq |x|^2 \tag{1}$$

We get $$|x|=(\sum_{i=1}^m |x_i|^2)^{\frac{1}{2}} \leq \sum_{i=1}^m |x_i| \tag{2}$$

By Cauchy inequality, $$\sum_{i=1}^m |x_i|=\sum_{i=1}^m 1 \cdot |x_i| \leq \sum_{i=1}^m 1^2 \cdot |x_i|^2)^\frac{1}{2} \leq \sqrt{m}(\sum_{i=1}^m (x_i)^2)^\frac{1}{2}=\sqrt{m} |x|$$, hence proved.

Can someone explain the tags 1 and 2 please?

In the first it means the following. $$|x|=\left|\sum_{i=1}^mx_i\right|=\sqrt{\left(\sum_{i=1}^mx_i\right)^2}=\sqrt{\sum_{i=1}^mx_i^2+2\sum_{1\leq i $$\leq\sqrt{\sum_{i=1}^mx_i^2+2\sum_{1\leq i I think it's better to see it by the triangle inequality.
The left inequality is wrong. Try $$x=0$$ and $$x_1=1$$.