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This question already has an answer here:

Show for any real numbers $a_1,a_2,...,a_n$
$$(a_1+a_2+···+a_n)^2 \leq n(a_1^2+a_2^2+...+a_n^2)$$

I know the definition of Cauchy-Schwarz is $$(\sum_{i=1}^n a_ib_i)^2 \leq \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2$$ I can write the same problem with following form, but I would not know how to continue to prove it $$(\sum_{i=1}^n a_i)^2 \leq n(\sum_{i=1}^n a_i^2)$$

Note: an other definition of Cauchy Schwarz $$ \langle\ v,u\rangle^2 \leq \langle\ u,u\rangle \langle\ v,v\rangle$$

Could you give me a steps what to use?

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marked as duplicate by Arnaud D., Community Sep 20 '18 at 21:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint:

Choose $b_i$ wisely. It is a constant.

If it is not clear which constant to pick, just let it be an arbitrary non-zero constant.

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Your statement doesn't look correct. Let n be 1 and $a_1$ be 1/2. Then the statement would say $1/2 \leq (1/2)^2$, which isn't true.

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    $\begingroup$ did you left out the square on the LHS? $\endgroup$ – Siong Thye Goh May 17 '18 at 2:40
  • $\begingroup$ I think that's new; an update since I originally posted. I'm on my phone though, so i could be wrong. I may have missed it. @Thye Goh $\endgroup$ – NicNic8 May 17 '18 at 4:19
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I get an other solution using the second definition of Cauchy-Schwarz $$ \langle u,v \rangle \leq \langle u,u \rangle \langle v,v \rangle $$

take $u = (1,1,...,1)$ as an vector of order n and be $v = (a_1,a_2,...,a_n)$

Then $ \langle u,v \rangle = (a_1+a_2+...+a_n) $

$ \langle u,u \rangle = \sqrt{n}$

$ \langle v,v \rangle = \sqrt{a_1^2+a_2^2+...+a_n^2}$

If remove the square root we have: $$ (a_1+a_2+...+a_n)^2 \leq n(a_1^2+a_2^2+...+a_n^2) $$

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