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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.

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

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    $\begingroup$ Second "<" in the penultimate line of your proof, hmm, I don't catch it: Could you hint at how to get to factor "6" et al? $\endgroup$ – Hanno Feb 15 '17 at 7:44
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    $\begingroup$ @Hanno and Cherng-tiao Perng: The second inequality in the last derivation is wrong. Both sides are geometric series the values of which violate the inequality. $\endgroup$ – Hans Jan 31 '18 at 23:20

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