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How can I prove that $\sum_{i=1}^{n} |a_i| \leq \sqrt{n} \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$ considering that $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers?

This exercise was presented in a section that also covered Cauchy Schwarz: $ (\sum_{i=1}^{n}a_{i}b_{i})^2 \leq (\sum_{i=1}^{n}a_{i}^2)(\sum_{i=1}^{n}b_{i}^2) $ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.

On the RHS, I proceeded with $ \sqrt{n} \sqrt{\sum_{i=1}^{n} a_{i}^{2}} $ = $ \sqrt{n\sum_{i=1}^{n} a_{i}^{2} } $ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.

Thank you!

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    $\begingroup$ Write $|a_i| = 1\cdot |a_i|$ and apply C-S. $\endgroup$ – Song Jan 11 at 21:44
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Let $b_i =1$ for $1\le i\le n$

Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$

You get the result right away upon taking square root of both sides.

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It's $$\sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}\geq$$ $$\geq\sqrt{1^2\cdot a_1^2}+\sqrt{1^2\cdot a_2^2}+...\sqrt{1^2\cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$

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