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As the title says, prove that $ \sqrt{\sum_{j=1}^{n}a_j^2} \le \sum_{j=1}^{n}|a_j|$, for all $a_i \in \mathbb{R}, i=\{1,2,...,n\}$. I can solve for the case of n=2, but kind of stuck while proving the n-terms version. The attempt is as follow:

$\left(\sqrt{\sum_{j=1}^{n}a_j^2}\right)^2 = \sum_{j=1}^n {a_j}^2 \le |\sum_{j=1}^n {a_j}^2| \le \sum_{j=1}^n |a_j|^2 \mbox{by triangle inequality.}$

Alright, I am stuck in the last inequality. Since I already square at the beginning, I would like to obtain something of the form squared too for the last inequality so that I can take the square root. When I solve the case for n=2, I can easily add terms to turn it into a square, but for this I am stuck. Help is pretty much appreciated.

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Equivalently, you want to show that $\left(\sum|a_j|\right)^2 \ge \sum a_j^2$.

If you expand $\left(\sum|a_j|\right)^2$, you will get $\sum a_j^2$ plus "cross" terms which are non-negative. More precisely, $$\left(\sum|a_j|\right)^2=\sum a_j^2 +2\sum_{i\lt j}|a_i||a_j|.$$

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  • $\begingroup$ To write down fully the last inequality $$\sum_{j=1}^n |a_j|^2 \le \sum_{j=1}^{n} |a_j|^2 + \sum \sum |a_i||a_j| \quad \mbox{index} \quad i \not= j $$ ? $\endgroup$ – Daniel Jun 25 '12 at 22:12
  • $\begingroup$ Thanks a lot! Just realized your edit after I wrote down. $\endgroup$ – Daniel Jun 25 '12 at 22:16

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