Bounding double sum off diagonal

Question

Let $\{a_n\}_{n\in\mathbb{N}}\subset\mathbb{C}$ and $N\in\mathbb{N}$. I have to prove the following bound $$ \sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}}|a_ma_n|\left(\log\frac{m}{n}\right)^{-2}\leq\sum_{n\leq N}|a_n|^2\sum_{1\leq m <n}\left(\frac{m}{n-m}\right)^2 $$ I think I have to use the fact that $1-\frac{1}{x}\leq \log x$ to get $$ \left(\log \frac{m}{n}\right)^{-2}\leq\left(\frac{m}{n-m}\right)^2 $$ but then I don't know how to reduce the term off the diagonal to a sum only of the diagonal terms. Any advice is welcome, thank you!

Edit. The above inequality is not true for $N=2$ as pointed out by Doyun Nam. What I eventually wanted to show is the following bound $$ \sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}}|a_ma_n|\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2|a_n|^2 $$ Maybe is easier to directly prove such bound without pasing first through the first inequality that I have written (which turned out to be wrong).

Answer 1

For each $n$ put $b_n=|a_n|$ . Then $\{b_n\}$ is a sequence of non-negative real numbers and we have to show that
$$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}} b_mb_n\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2b_n^2 $$

$$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}} b_mb_n\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2b_n^2$$ $$\sum_{n\leq N}\sum_{\substack{m\leq N }} b_mb_n \left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}(n^2+1)b_n^2 $$ $$\left( \sum_{n\leq N} b_nn^2\right)\left(\sum_{\substack{m\leq N }} b_mm^{-2}\right)\ll \sum_{n\leq N}(n^2+1)b_n^2 $$ Clearly, this inequality holds asymptotically iff $\sum_{m=1}^\infty b_mm^{-2}<\infty$.