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A simpler version of Hilbert's Inequality states that: For any real numbers $a_1,a_2\cdots,a_n$ the following inequality holds: $\sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{i+j}\leq\pi\sum_{i=1}^na_i^2$.

I was reading a proof of this inequality where first they applied Cauchy Schwarz to get $(\sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{i+j})^2\leq(\sum_{i=1}^n\sum_{j=1}^n\frac{\sqrt{i}a_i^2}{\sqrt{j}(i+j)})(\sum_{i=1}^n\sum_{j=1}^n\frac{\sqrt{j}a_j^2}{\sqrt{i}(i+j)})$.

Then they stated that it suffices to prove $\sum_{n=1}^{\infty}\frac{\sqrt{m}}{(m+n)\sqrt{n}}\leq\pi$ for any positive integer $m$.

Can someone explain why this is true? I tried manipulating the expression in a lot of different ways but couldn't conclude the result above. I'd appreciate any ideas or thoughts.

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  • $\begingroup$ Did you get the direction of the inequality wrong? $\endgroup$
    – Hans
    Aug 18, 2015 at 21:20
  • $\begingroup$ Yeah oops, but it's fixed it now. $\endgroup$
    – Max
    Aug 18, 2015 at 21:22

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You can show that $\displaystyle\sum_{n=1}^\infty\,\frac{\sqrt{m}}{(m+n)\sqrt{n}}<\int_0^\infty\,\frac{\sqrt{m}}{(m+x)\sqrt{x}}\,\text{d}x=\Bigg.\Bigg(2\arctan\left(\sqrt{\frac{x}{m}}\right)\Bigg)\Bigg|_{x=0}^{x=\infty}=\pi$.

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  • $\begingroup$ Yes, that follows from the fact that $\frac{\sqrt{m}}{(m+n)\sqrt{n}}$ is monotonically increasing. But why does it suffice to prove $\sum_{n=1}^{\infty} \frac{\sqrt{m}}{(m+n)\sqrt{n}}\leq\pi$? $\endgroup$
    – Max
    Aug 18, 2015 at 21:27
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    $\begingroup$ That is because $$\sum_{i=1}^n\, \sum_{j=1}^n\frac{\sqrt{i}a_i^2}{\sqrt{j}(i+j)} =\sum_{i=1}^n\,a_i^2\,\sum_{j=1}^n\frac{\sqrt{i}}{\sqrt{j}(i+j)}=\sum_{i=1}^n\, \sum_{j=1}^n\frac{\sqrt{j}a_j^2}{\sqrt{i}(i+j)}\,.$$ $\endgroup$
    – Batominovski
    Aug 18, 2015 at 21:31

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