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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?

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    $\begingroup$ Can you please add the self-study tag and read its wiki? $\endgroup$ – kjetil b halvorsen Oct 30 '18 at 22:53
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    $\begingroup$ 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 ($P$). $\endgroup$ – whuber Oct 31 '18 at 3:03
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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

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

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

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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.

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