# 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 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_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);
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}{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)=
frac{pi}2left(sqrt{AB}+sqrt{AB}right)=pisqrt{AB}.\$\$
Squaring both sides and replacing \$A\$ and \$B\$, we obtain the desired inequality