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I need to show that:

$ {\sum\limits_{i=1}^n {|x|} } \leq \sqrt{n\sum\limits_{i=1}^n |x|^2 } $

I tried to square both sides so I would get:

$ \left({\sum\limits_{i=1}^n {|x|} }\right)^2 = \left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i j}|x_i||x_j|\right) \leq n\sum\limits_{i=1}^n |x|^2 $

but it just doesn't seem to work...

I know that on both sides we have $n^2$ elements, I just don't know how to compare them.

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    @AndréNicolas Thanks, re-posted with a little explication. You should be credited with recognizing the typo or mis-transcription!2012-11-27

2 Answers 2

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Make sure you don't have a typo, and that you copied the question correctly. I suspect you need to work with the following:

$\left(\sum_{i=1}^n|x_i|\right)^2 \le n \sum_{i=1}^n|x_i|^2.$

Now, each side of the inequality has $n^2$ terms.

You can use the Cauchy-Schwarz Inequality. As applied to Euclidean space $\mathbb{R}^n$:

$\left(\sum_{i=1}^n x_iy_i\right)^2 \le \left(\sum_{i=1}^nx_i^2 \right)\left(\sum_{i=1}^n y_i^2 \right) $

For your problem, $\left(\sum_{i=1}^n x_iy_i\right)^2 = \left(\sum_{i=1}^n x_i\cdot 1\right)^2$


Alternatively, if you have the following inequality to prove: $\left(\sum_{i=1}^n|x_i|\right) \le \sqrt{n \sum_{i=1}^n|x_i|^2}.$

Then simply square both sides of this inequality to obtain the inequality at the top, and proceed as suggested.

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    nicely answered! +12013-05-23
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Because $x^2$ is a convex function, Jensen's Inequality says $ \left(\frac1n\sum_{k=1}^nx_k\right)^2\le\frac1n\sum_{k=1}^nx_k^2 $ Multiplying by $n^2$ yields $ \left(\sum_{k=1}^nx_k\right)^2\le n\sum_{k=1}^nx_k^2 $ Taking the square root gives $ \sum_{k=1}^nx_k\le\sqrt{n\sum_{k=1}^nx_k^2} $