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It seems that the following functions $$G(R,s)=(1-R^2)\int_0^1\int_0^{2\pi} Adr da,$$ where $$A=\frac{ \sqrt{\left(\left(1+r^2\right) \cos(a+s)-2 r R \cos s\right)^2+\left(1-r^2\right)^2 \sin^2(a+s)}}{\left(1+r^2 R^2-2 r R \cos a\right)^2}$$ attains its maximum for $R=0$, if $R\in[0,1]$, however I dont have the proof.

Geometricaly, it is equivalent to find $$\sup_{z\in U}\int_{U}\left|\frac{1}{z-w}+\frac{w}{1-\bar z w}\right|du dv, \ \ w=u+iv,$$ where $U$ is the unit disk.

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    $\begingroup$ This integral seems to have some geometric meaning. If you have such, would you please share it? $\endgroup$
    – Fedor Petrov
    Commented Jul 16, 2020 at 18:04
  • $\begingroup$ @Fedor Petro: I made some geometric explanation. $\endgroup$
    – Lira
    Commented Jul 16, 2020 at 18:26
  • $\begingroup$ @ Fedor Petrov. Yes, $w=u+iv$. $\endgroup$
    – Lira
    Commented Jul 16, 2020 at 18:37
  • $\begingroup$ Written as $\sup_{z\in U}\int_{U}\left|\frac{1}{z-w}-\frac{1}{\bar z -1/w}\right|du dv$, it looks like the integral only depends on |z| and it might just follow from some symmetry considerations that the maximum is attained at z=0. $\endgroup$
    – Wolfgang
    Commented Jul 16, 2020 at 20:26
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    $\begingroup$ I don't see why the two expressions you give for $G(R,s)$ have the same values. If you write $w=\frac rR\,ze^{ia}$ with $R:=|z|$ and $r:=|w|$, then $|1-\bar zw|^2=1+r^2R^2-2rR\cos a$; but how come the denominator of the expression for $A$ is $(1+r^2R^2-2rR\cos a)^2$? $\endgroup$
    – Iosif Pinelis
    Commented Jul 16, 2020 at 23:31

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