Can some one help to prove $\newcommand{\Z}{\mathbb{Z}}$ $$ \sum_{k\in\Z,k\not=n} \sum_{l\in\Z,l\not = n} \frac{\langle k-n\rangle }{\langle k+l\rangle |k-n|\langle l + n\rangle |l-n|} \langle k+l\rangle |f_{k+l}| \langle l\rangle |g_l| \\ \leq \left( \sum_{k\in\Z,k\not=n} \sum_{l\in\Z,l\not = n} \frac{\langle k-n\rangle^2 }{\langle k+l\rangle^2 |k-n|^2\langle l + n\rangle^2 |l-n|^2} \right)^{1/2}\left(\sum_{k\in\Z}\langle k\rangle^2 |f_k|^2 \right)^{1/2} \left(\sum_{k\in\Z}\langle k\rangle^2 |g_k|^2 \right)^{1/2}, $$ where $\langle n\rangle = 1+|n|$.
It is supposed to be a consequence of Cauchy-Schwarz and Young's inequality for convolutions but I cant manage to prove it.
Is it that easy and I'm just not able or is this a known result? It looks like a weighted convolution inequality, is this true if I replace the fraction with any kernel $K(k,l)$?
By Cauchy-Schwarz inequality we have \begin{align} &\ \sum_{k, l \neq n} \frac{\langle k-n\rangle}{\langle k+l\rangle |k-n| \langle n+l \rangle |n-l| } \langle k+l\rangle |f_{k+l}|\langle l\rangle |g_l|\\ \leq&\ \left(\sum_{k, l \neq n} \frac{\langle k-n\rangle^2}{\langle k+l\rangle^2 |k-n|^2 \langle n+l \rangle^2 |n-l|^2 }\right)^{1/2} \left(\sum_{k, l \neq n} \langle k+l\rangle^2|f_{k+l}|^2\langle l\rangle^2|g_l|^2\right)^{1/2}\\ =&\ \left(\sum_{k, l \neq n} \frac{\langle k-n\rangle^2}{\langle k+l\rangle^2 |k-n|^2 \langle n+l \rangle^2 |n-l|^2 }\right)^{1/2} \left(\sum_{l \neq n}\left[\sum_{ k \neq n} \langle k+l\rangle^2|f_{k+l}|^2\right]\langle l\rangle^2|g_l|^2\right)^{1/2}\\ =&\ \left(\sum_{k, l \neq n} \frac{\langle k-n\rangle^2}{\langle k+l\rangle^2 |k-n|^2 \langle n+l \rangle^2 |n-l|^2 }\right)^{1/2} \left[\sum_{l \neq n}\langle l\rangle^2|g_l|^2\right]^{1/2}\left[\sum_{ k \neq n} \langle k\rangle^2|f_{k}|^2\right]^{1/2}. \end{align}