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In my previous MO question, the inequality was about a specific series and nicely answered by Cherng-tiao Perng. After testing with a few more numerical infinite sums, I came to realize that perhaps more is true.

Does the following hold true for any convergent series? $$\left(\sum_{k\geq1}t_k\right)^4 <\pi^2\left(\sum_{k\geq1}t_k^2\right)\left(\sum_{k\geq1}k^2t_k^2\right).$$ Is the constant $\pi^2$ optimal?

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This is known to be Carlson's inequality from 1935 (for $t_k\geq 0$, and not all $t_k$ are $0$). The Swedish mathematician Fritz Carlson (1888-1952) also proved the optimality of the constant $\pi^2$. For an elegant elementary proof of the inequality see G. H. Hardy, A note on two inequalities, J. London Math. Soc. 11, 167-170, 1936 (DOI https://doi.org/10.1112/jlms/s1-11.3.167 or http://onlinelibrary.wiley.com/doi/10.1112/jlms/s1-11.3.167/abstract).

Note: This inequality has found many modern day applications and generalizations, see the book "Multiplicative Inequalities of Carlson Type and Interpolation" by L. Larsson et al., World Scientific, 2006. (DOI https://doi.org/10.1142/6063; there you will find a free sample chapter with the classical proofs of Carlson and Hardy.)

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Following Cherng-tiao Perng's proof for the other inequality, here is what I find. I don't have immediate access to Folkmar Bornemann's references, so I'm not sure the ideas here might be similar. No originality is claimed.

Update. A short visit to the library confirms that Hardy's paper contains the below proof, only details are added for the reader's convenience.

Begin by writing $t_k=\frac1{\sqrt{\alpha+\beta k^2}}t_k\sqrt{\alpha+\beta k^2}$, for $\alpha, \beta>0$ to be specified later. Then, we apply the Cauchy-Schwarz inequality as follows \begin{align}\left(\sum_kt_k\right)^2&=\left(\sum_k\frac1{\sqrt{\alpha+\beta k^2}}t_k\sqrt{\alpha+\beta k^2}\right)^2 \leq\left(\sum_{k=1}^{\infty}\frac1{\alpha+\beta k^2}\right)\left(\sum_{k=1}^{\infty}(\alpha+\beta k^2)t_k^2\right)\\ &<\left(\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}\right)\left(\alpha\sum_{k=1}^{\infty}t_k^2+\beta\sum_{k=1}^{\infty}k^2t_k^2\right) =\frac{\pi}{2\sqrt{\alpha\beta}}\left(\alpha A+\beta B\right); \end{align} where we denoted $A:=\sum_kt_k^2$ and $B:=\sum_kk^2t_k^2$. Of course, $\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}=\frac{\pi}{2\sqrt{\alpha\beta}}$ is from Calculus. At this stage, we make a convenient choice of $\alpha=\sqrt{\frac{B}A}$ and $\beta=\sqrt{\frac{A}B}$. Clearly, $\alpha\beta=1$. So, $$\left(\sum_kt_k\right)^2<\frac{\pi}{2\sqrt{\alpha\beta}}\left(\alpha A+\beta B\right)=\frac{\pi}2\left(\sqrt{\frac{B}A}A+\sqrt{\frac{A}B}B\right)= \frac{\pi}2\left(\sqrt{AB}+\sqrt{AB}\right)=\pi\sqrt{AB}.$$ Squaring both sides and replacing $A$ and $B$, we obtain the desired inequality $$\left(\sum_kt_k\right)^4<\pi^2AB=\pi^2\sum_kt_k^2\sum_kk^2t_k^2.\qquad \square$$

Remark. $\sum_{k=1}^{\infty}\frac1{\alpha+\beta k^2}<\int_0^{\infty}\frac{dx}{\alpha+\beta x^2}$ is due to Lower Riemann Sums with partition $\{0,1,2,3,4,\dots\}$.

Remark. I've to find Hardy's paper to see why $\pi^2$ is optimal.

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    $\begingroup$ Gorgeous proof. Will have to read through more carefully later. $\endgroup$ – Brevan Ellefsen Feb 12 '17 at 19:56
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    $\begingroup$ This is exactly Hardy's proof, except for minor some variations of notation, as everyone can see by following the link given above (that is to say, the first page of Hardy's 1936 paper is shown directly at the webpage and gives all what is stated here). $\endgroup$ – Folkmar Bornemann Feb 12 '17 at 20:36
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    $\begingroup$ When I follow the link above, I get a page that says "We're sorry! Your action has resulted in an error. Please click the Back button in your browser and try again." So I'm not sure it's true that everyone can see it. $\endgroup$ – Gareth McCaughan Feb 13 '17 at 13:10
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    $\begingroup$ The DOI link seems to be (temporarily) broken. Use onlinelibrary.wiley.com/doi/10.1112/jlms/s1-11.3.167/abstract instead, you will also get a preview of the first page of Hardy's 1936 paper. $\endgroup$ – Folkmar Bornemann Feb 13 '17 at 14:30

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