
> 
> **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.



seeking proofs: infinite series inequalities
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.


