Is $\frac{a_1+\cdots+a_n}{\sqrt{n(b_1+\cdots+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$?

Question

For $a_1>0$, $a_2>0,\dots,a_n>0$, and $b_1>0$, $b_2>0,\dots,b_n>0$

I want to prove: $$\frac{a_1+a_2+\dots+a_n}{\sqrt{n(b_1+b_2+...+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$$

Answer 1

For $n=2$ your inequality reduces to

$$\frac{a_1+a_2}{\sqrt{2(b_1+b_2)}} \leq \frac{1}{2} (\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}})$$

Lets the $a_1=1,a_2=\frac{1}{2},b_1=1,b_2=\frac{11}{16}$ then you have

$$\sqrt{\frac{2}{3}} \leq \frac{1}{2}+\frac{1}{\sqrt{11}}$$

This is false, so the inequality doesn't hold.

Answer 2

This actually isn't true. Let $r_i = {a_i \over \sum_i a_i}$. Then what you are trying to prove is equivalent to $${1 \over \sqrt{\sum_{i=1}^nb_i \over n}} \leq \sum_{i=1}^n \bigg(r_i {1 \over \sqrt{b_i}}\bigg)$$ Here $\sum_i r_i = 1$ and each $r_i > 0$. But notice that $\sum_{i=1}^n{b_i \over n}$ is the average of the $b_i$. So you are trying to show that $${1 \over \sqrt{b_{average}}} \leq \sum_{i=1}^n \bigg(r_i {1 \over \sqrt{b_i}}\bigg)$$ All you have to do is make the $r_i$ for the largest $b_i$ nearly one, and the others nearly zero, and the inequality won't hold.

So for any choice of $b_i$'s that aren't all the same, you can find some $a_i$'s for which this doesn't work.