Is it true that the following weighted-average is smaller than the respective arithmetic average
$$
\sum_{n=1}^{N}\left(\frac{b_{n}}{\sum_{m=1}^{N}b_{m}}\right)\frac{a_{n}}{b_{n}}\overset{?}{\le}\frac{1}{N}\sum_{n=1}^{N}\frac{a_{n}}{b_{n}}
$$
for all $a_n \in [0,1]$, $b_n \in [1, D]$, and $D \ge 2$?
Without the $a_n$'s it would simply be true by Jensen's inequality and the convexity of $x\mapsto 1/x$ (for positive $x$'s), but with the $a_n$ in the mix it is not so clear and it seems one cannot directly apply Chebyshev's sum inequality either.
Edit: it is easy to see that one can upper bound the numerator in the parenthesis by $D$ and lower bound the corresponding denominator by $N$, obtaining the desired upper bound with an extra $D$ term. I was wondering however if one can do better in general.
Best Answer
Testing the claim for $N=2$, suppose $b_2 > b_1$. \begin{align*} \text{Then}\;\, & \left(\frac{b_1}{b_1+b_2}\right)\left(\frac{a_1}{b_1}\right) + \left(\frac{b_2}{b_1+b_2}\right)\left(\frac{a_2}{b_2}\right) \le \frac{1}{2}\left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right)\\[4pt] \iff\;& \frac{a_1+a_2}{b_1+b_2}\le\frac{a_1b_2+a_2b_1}{2b_1b_2}\\[4pt] \iff\;& (b_1+b_2)(a_1b_2+a_2b_1)-(2b_1b_2)(a_1+a_2)\ge 0\\[4pt] \iff\;& (b_2-b_1)(a_1b_2-a_2b_1)\ge 0\\[4pt] \iff\;&a_1b_2-a_2b_1\ge 0\\[4pt] \iff\;&a_1b_2\ge a_2b_1\\[4pt] \iff\;&\frac{a_1}{a_2}\ge \frac{b_1}{b_2}\\[4pt] \end{align*} which need not be true.