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

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.
