An attempt to generalize the previous inequality

This is known to be Carlson’s inequality from 1935 (for $t_kgeq 0$, 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 or

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; 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 $t_k=frac1{sqrt{alpha+beta k^2}}t_ksqrt{alpha+beta k^2}$, for $alpha, beta>0$ to be specified later. Then, we apply the Cauchy-Schwarz inequality as follows
begin{align}left(sum_kt_kright)^2&=left(sum_kfrac1{sqrt{alpha+beta k^2}}t_ksqrt{alpha+beta k^2}right)^2
leqleft(sum_{k=1}^{infty}frac1{alpha+beta k^2}right)left(sum_{k=1}^{infty}(alpha+beta k^2)t_k^2right)\
&<left(int_0^{infty}frac{dx}{alpha+beta x^2}right)left(alphasum_{k=1}^{infty}t_k^2+betasum_{k=1}^{infty}k^2t_k^2right)
=frac{pi}{2sqrt{alphabeta}}left(alpha A+beta Bright);
where we denoted $A:=sum_kt_k^2$ and $B:=sum_kk^2t_k^2$. Of course, $int_0^{infty}frac{dx}{alpha+beta x^2}=frac{pi}{2sqrt{alphabeta}}$ is from Calculus. At this stage, we make a convenient choice of $alpha=sqrt{frac{B}A}$ and $beta=sqrt{frac{A}B}$. Clearly, $alphabeta=1$. So,
$$left(sum_kt_kright)^2<frac{pi}{2sqrt{alphabeta}}left(alpha A+beta Bright)=frac{pi}2left(sqrt{frac{B}A}A+sqrt{frac{A}B}Bright)=
Squaring both sides and replacing $A$ and $B$, we obtain the desired inequality
$$left(sum_kt_kright)^4<pi^2AB=pi^2sum_kt_k^2sum_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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