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.