# Summation of squared x_i if summation of x_i is 1

How to prove "If $$\sum_{i=1}^n x_i=1$$, then $$\sum_{i=1}^n x_i^2>1/n$$"? I'm thinking about $$Var(x_i)=E(x_i^2)-[E(x_i)]^2=\frac{1}{n}\sum_{i=1}^n x_i^2-1/n^2\ge0$$. Is that correct?

• Can you please add the self-study tag and read its wiki? – kjetil b halvorsen Oct 30 '18 at 22:53
• There is a pleasant geometrical interpretation: No point on the hyperplane passing through the point $P=(1/n, 1/n,\ldots, 1/n)$ with normal direction $nP=(1,1,\ldots, 1)$ lies in the interior of the origin-centered ball of squared radius $|P|^2 = 1/n^2 + \cdots 1/n^2 = 1/n.$ The proof is that this hyperplane is (obviously) tangent to the boundary of that ball, because its normal vector is parallel to the radius vector at one point of intersection ($P$). – whuber Oct 31 '18 at 3:03

Try writing variance as $$Var(x_i)=E(x^2_i)−[E(x_i)]^2$$

substitute $$[E(x_i)]^2 = 1/n^2$$

$$1/n\Sigma (x_i - \bar x)^2 = 1/n\Sigma x_i^2 - 1/n^2$$

from there

$$1/n\Sigma x_i^2 = 1/n\Sigma(x_i - \bar x)^2 + 1/n^2)$$

$$\Sigma x_i^2 = (\Sigma(x_i - \bar x)^2 + 1/n)$$

which is the same as

$$\Sigma x_i^2 = (\Sigma(x_i - \bar x)^2 + 1/n)$$

or $$\Sigma x_i^2 = n*Var(x_i) + 1/n$$

Since variance and n are greater than 0

$$n*Var(x_i) + 1/n > 1/n$$

$$\Sigma x_i^2 > 1/n$$

Hint: Can you calculate the (arithmetic) mean of the $$x_i$$, $$\bar{x}=\frac1{n}\sum_{i=1}^n x_i$$? Then write $$\sum_{i=1}^n x_i^2 = \sum_{i=1}^n\left( (x_i-\bar{x})+ \bar{x}\right)^2$$ and you should be able to conclude.