Question. Numerically, the following is convincing. However, is there a proof? $$\left(\sum_{k\geq1}\frac1{\sqrt{2^k+3^k}}\right)^4 <\pi^2\left(\sum_{k\geq1}\frac1{2^k+3^k}\right)\left(\sum_{k\geq1}\frac{k^2}{2^k+3^k}\right).$$
This comes up in some recent work and the inequality seems needed.
This may serve as a different approach. By Cauchy-Schwarz inequality, $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \left(\sum_{k\geq 1}\frac 1{k^2}\right)\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right),$$ which shows that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \frac{\pi^2}6\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right).$$
So it suffices to show that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2<6\cdot \sum_{k\geq 1}\frac 1 {2^k+3^k}.$$
But \begin{align} \left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2 &<\left(\sum_{k\geq 1}\frac 1{\sqrt{2\cdot 6^{k/2}}}\right)^2 \\ &<6\left(\frac 1{2+3}+\sum_{k\geq 2}\frac 1{2\cdot 3^k}\right) \\ &<6\left(\sum_{k\geq 1}\frac 1{2^k+3^k}\right), \end{align} where in the first inequality, the AM-GM inequality was used.
