An attempt to generalize the previous inequality

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.)

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.