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This question already has an answer here:

I want to prove that for all $n\in\mathbb{N}$ and $x_i\in\mathbb{R}$, we have $$(x_1+\cdots+x_n)^2\leq n(x_1^2+\cdots+x_n^2).$$

At first I wanted to do a proof by induction, but it has turned out to be much more complicated than I imagined. Now I'm thinking Cauchy-Schwarz, but I can't seem to write it in the correct form to get the result I need.

Any hints or help would be much appreciated. The Cauchy-Schwarz inequality as we are using it in my class is in vector form, i.e., $$|\langle u,v\rangle|\le\|u\| \|v\|.$$

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marked as duplicate by Yuval Filmus, drhab, Tom-Tom, TZakrevskiy, Daniel W. Farlow Oct 27 '15 at 16:10

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  • $\begingroup$ It's enough to prove it for positive reals ($|x|\ge x$). Use Jensen's inequality. $f(x)=x^2$ is concave for positive $x$, so $$f\left(\frac{x_1+x_2+\cdots+ x_n}{n}\right)\le \frac{f(x_1)+f(x_2)+\cdots f(x_n)}{n}$$ $\endgroup$ – user236182 Oct 27 '15 at 2:50
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$$ \left( \sum_{k=1}^{n}x_{k} \right)^{2} = \left( \sum_{k=1}^{n}x_{k}\cdot 1 \right)^{2} \leq \left (\sum_{k=1}^{n}x_{k}^{2} \right) \left( \sum_{k=1}^{n}1 \right) = n \sum_{k=1}^{n}x_{k}^{2}. $$

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You can also solve it with Chebyshev's Inequality (which is easy to prove, as shown in the link).

Wlog let $x_1\ge x_2\ge\ldots\ge x_n$. Then $$n\left(\sum_{i=1}^n x_ix_i\right)\ge \left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n x_i\right)$$

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You can get a direct proof by noting that $$ \sum_{1 \leq i < j \leq n} (x_i - x_j)^2 = (n-1) \sum_{1 \leq i \leq n} x_i^2 - 2 \sum_{1 \leq i < j \leq n} x_i x_j = n \sum_{1 \leq i \leq n} x_i^2 - \left(\sum_{1 \leq i \leq n} x_i\right)^2. $$


Another possibility is given a probabilistic interpretation of user236182's proof. Let $X$ be a random variable ranging uniformly on the multiset $\{x_1,\ldots,x_n\}$. The variance of $X$ is $$ V[X] = E[X^2] - E^2[X] = \frac{1}{n} \sum_{i=1}^n x_i - \frac{1}{n^2} \left(\sum_{i=1}^n x_i \right)^2. $$


Yet another possibility is to minimize the right-hand side while keeping the left-hand side fixed. Suppose that $y + \delta \leq z - \delta$ for some $\delta \geq 0$. Then $$ (y + \delta)^2 + (z - \delta)^2 = y^2 + z^2 + 2\delta (y - z + \delta) \leq y^2 + z^2. $$ Let $\mu$ be the average of $x_1,\ldots,x_n$. If the vector $x_1,\ldots,x_n$ is not constant then we can always find some $i,j$ such that $x_i < \mu < x_j$. If $\mu \leq (x_i + x_j)/2$ then $x_i + (\mu - x_i) \leq x_j - (\mu - x_i)$, and so if we replace $x_i,x_j$ by $\mu,x_i+x_j-\mu$, not changing the left-hand side and not increasing the right-hand side. The case $\mu > (x_i + x_j)/2$ can be handled similarly. As a result, we have changed the values of $x_1,\ldots,x_n$ in a way that doesn't change the left-hand side and doesn't increase the right-hand side, and in addition increases the number of $x_i$s equal to $\mu$.

Repeating this operation at most $n-1$ times, we reach the constant $\mu$ vector, for which the inequality trivially holds (both sides equal $(n\mu)^2$). Looking back at the original inequality, its right-hand side cannot be any smaller than $(n\mu)^2$ (since our modification can't have increased it), and we deduce the inequality.

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  • $\begingroup$ I really like the second proof, +1. $\endgroup$ – wythagoras Oct 26 '15 at 17:35

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