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Prove that, for any sequences of real numbers $\{a_n\}$ and $\{b_n\}$, we have $$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{a_mb_n}{m+n}\le \pi\left(\sum_{m=1}^\infty a_m^2\right)^{1/2}\left(\sum_{n=1}^\infty b_n^2\right)^{1/2} $$

How can I solve this problem? I've got no idea for it... Please help!

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    $\begingroup$ Hint: $\displaystyle\pi=\sqrt6\cdot\bigg[\sum_{k=1}^\infty\frac1{k^2}\bigg]^{1/2}-$ See Basel problem for more information. Then use the Cauchy product, along with some inequalities related to quadratic and geometric means. $\endgroup$ – Lucian May 17 '14 at 6:25
  • $\begingroup$ Would you clarify the usage of those things? It's seems awkward :( $\endgroup$ – La classe May 17 '14 at 8:45
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By using Cauchy-Schwarz inequality in the following form: $$\left(\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{a_mb_n}{m+n} \right)^2\le \left(\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{a_m^2}{m+n}\sqrt{\frac{m}n} \right) \left(\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{b_n^2}{m+n}\sqrt{\frac{n}m} \right) \tag {$\star$}$$

for the first factor on the RHS, we have $$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{a_m^2}{m+n}\sqrt{\frac{m}n} = \sum_{m=1}^\infty a_m^2 \sum_{n=1}^\infty\frac{\sqrt{m/n}}{m+n}$$

Further we have $$\sum_{n=1}^\infty\frac{\sqrt{m/n}}{m+n} \le \int_0^\infty \frac{\sqrt{m/x}}{m+x}dx = \int_0^\infty \frac{dt}{\sqrt t \;(1+t)} = \pi$$ where we have used $x = mt$ as a substitution. By symmetry the second factor in the RHS of $\star$ also can be treated similarly, giving you the result.

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