# How to prove $\left(\sum_{i=1}^n x_i\right)^2 \le n\sum_{i=1}^n x_i^2$?

For now I have the following \begin{align*} \left(\sum_{i=1}^n x_i\right)^2 & \le n\sum_{i=1}^n x_i^2 \\ \sum_{i=1}^n x_i^2 + 2\sum_{i<j}x_ix_j & \le n\sum_{i=1}^n x_i^2 \\ 2\sum_{i<j}x_ix_j & \le (n-1)\sum_{i=1}^n x_i^2 \\ \end{align*} Then I'm thinking about using Cauchy-Schwarz inequality. But I feel stuck.

• Use Cauchy-Schwartz immediately at the beginning. Find appropriate vectors and take the square root on the given inequality. – Git Gud Feb 10 '15 at 11:08

First let's recall the Cauchy - Schwarz inequality:

$$\left(\sum_{i=1}^n x_iy_{i}\right)^2 \le \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (y_{i})^2$$

Therefore by letting $y_{i} = 1$ for all $i$:

$$\left(\sum_{i=1}^n x_i \times 1\right)^2 \le \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (1)^2$$ $$\left(\sum_{i=1}^n x_i \right)^2 \le \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (1)$$ $$\left(\sum_{i=1}^n x_i \right)^2 \le \sum_{i=1}^n (x_i)^2 n$$ $$\left(\sum_{i=1}^n x_i \right)^2 \le n\sum_{i=1}^n (x_i)^2.$$

The Cauchy-Schwarz inequality tells you that $$\langle v,w\rangle^2\le\langle v,v\rangle\langle w,w\rangle$$ If you take $v=(x_1,x_2,\dots,x_n)$, you just need to find $w$ such that $$\langle v,w\rangle=x_1+x_2+\dots+x_n$$ and $\langle w,w\rangle=n$.

You can use $x_i^2+x_j^2\ge2x_ix_j$ (i.e. $(x_i-x_j)^2\ge0$) for all combinations $(i,j)$ with $j>i$ and sum the resulting inequalities. Every term $x_i^2$ occurs $(n-1)$ in the resulting sum.

For example, if n=3 $$x_1^2+x_2^2\ge 2x_1x_2 \\ x_1^2+x_3^2\ge 2x_1x_3 \\ x_2^2+x_3^2\ge 2x_2x_3$$