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I've been struggling with the following math problem:

Let $a_1,\cdots,a_n\in\mathbb{R}$

Show that $$\left(\frac 1n \sum_{k=1}^n a_k^2\right)^{\frac12} \leq \left(\frac 1n \sum_{k=1}^n a_k^4\right)^{\frac14}$$

I've been able to modify the equation into the following form, but I'm unable to proceed, since I don't know how to modify the summations:

$$\left(\frac 1n \sum_{k=1}^n b^k\right)^2\leq n\sum_{k=1}^nb_k^2$$

where $b_k$ is substituted for $a_k^2$

Could someone guide me through how to deal with those summations? Thanks a lot in advance!

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    $\begingroup$ Use Cauchy-Schwarz on your modified version. Apply it like $\sum a_k b_k$ where $a_k=1.$ $\endgroup$ Dec 20 '19 at 14:59
  • $\begingroup$ To write Maths equations on here, put a $ before and after them, or $$ to make them big and central. I have edited your post so you can see how it is done $\endgroup$
    – lioness99a
    Dec 20 '19 at 15:13
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Let $a_i^2=x_i$.

Thus, your inequality it's just Jensen for a convex function $f(x)=x^2:$ $$\frac{x_1^2+...+x_n^2}{n}\geq\left(\frac{x_1+...+x_n}{n}\right)^2.$$ Also, you can use C-S: $$\frac{x_1^2+...+x_n^2}{n}=\frac{1}{n^2}(1^2+...+1^2)(x_1^2+...+x_n^2)\geq\frac{1}{n^2}(x_1+...+x_n)^2.$$

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The second inequality you have it is just the fact that the variance of the $b_j$ is non negative.

Indeed, let $\overline{b}$ be the arithmetic mean of the $b_j$. The variance of $b_j$ is

$$0\le Var(b_j)=\frac1n \sum_j(\overline{b}-b_j)^2=\frac1n\sum_j(\overline{b}^2-2\overline{b}b_j+b_j^2)$$

Thus $$0\le\overline{b}^2-2\overline{b}\frac1n\sum_j b_j+\frac1n\sum_j b_j^2=\frac1n\left(\sum_j b_j^2\right)- \overline{b}^2$$

which is your thesis.

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