# Inequation involving summation

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!

• Use Cauchy-Schwarz on your modified version. Apply it like $\sum a_k b_k$ where $a_k=1.$ Dec 20 '19 at 14:59
• 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 Dec 20 '19 at 15:13

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: $$\frac{x_1^2+...+x_n^2}{n}=\frac{1}{n^2}(1^2+...+1^2)(x_1^2+...+x_n^2)\geq\frac{1}{n^2}(x_1+...+x_n)^2.$$

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