# Comparing sums with squares

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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You have not indicated the context, but the inequality as stated is not correct for the "natural" situation. Remove the square root fom the right side. Then we are dealing with the Cauchy-Schwarz Inequality with the $y_i=1$. – André Nicolas Nov 27 '12 at 17:46
@AndréNicolas Didn't see your comment until just a few moments ago! – amWhy Nov 27 '12 at 17:52
@amWhy: Why delete? If Cauchy-Schwarz was in fact my discovery, there might be a point. But in fact I am not that old. – André Nicolas Nov 27 '12 at 17:59
@AndréNicolas Thanks, re-posted with a little explication. You should be credited with recognizing the typo or mis-transcription! – amWhy Nov 27 '12 at 18:05

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! +1 – Amzoti May 23 '13 at 1:44

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 $$\sum_{k=1}^nx_k\le\sqrt{n\sum_{k=1}^nx_k^2}$$

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