# seeking proofs: infinite series inequalities

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.

• It seems you only need 3 terms for each sum on the right and a crude upper bound for the sum on the left. – Brendan McKay Feb 12 '17 at 7:27
• This is a research-level question? – user541686 Feb 13 '17 at 0:22

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 $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2<6\cdot \sum_{k\geq 1}\frac 1 {2^k+3^k}.$$