upper bound on logarithmic potential For $\tau \in [0,1)$, define the ellipse
$$E_{\tau} = \{ (x,y) : \frac{x^2}{(1+\tau)^2} + \frac{y^2}{(1-\tau)^2} \leq 1 \}\tag{1}$$
$\mu_{\tau}$ to be the uniform distribution on $E_{\tau}$ (i.e. $\mu_{\tau} = \frac{1}{\pi (1-\tau^2)} dx dy)$ and the function:
$$J(x,y) = \int_{E_{\tau}} \log |x+iy - z| d\mu_{\tau}(z) - \frac{x^2}{2(1+\tau)} - \frac{y^2}{2(1-\tau)}\tag{2}$$
I want to show that for any $x > 1+\tau$ and $y \in \mathbb{R}$,
$$J(x,y) \leq J(x,0)\tag{3}$$
My thoughts:


*

*By constructions, $J(x, \cdot)$ is an even function, so it suffices to only consider $y > 0$. For very large $y$, we have logarithmic asymptotics on the first term, so the inequalities follows simply. However, my main issue is getting near $y = 0$.

*For $\tau = 0$, we can explicitly compute $J$, since $E_{\tau}$ is just a disk at the point. In fact,
$$J(x,y) = \frac{1}{2}\log (x^2 + y^2) - \frac{1}{2}(x^2+y^2)\tag{4}$$
from which the inequality can be verified instantly.

*We can explicitly compute $J(x,0)$. The trick is to different $J$ as a function of $z$, which yields its Cauchy transform. In fact,
$$\partial_z J(z) = \frac{1}{2 \tau}(z - \sqrt{z^2 - 4\tau})\tag{5}$$
From here, if we restrict to real $z$, then we can simply integrate this and obtained an explicit formula. In general, however, I suppose we have to do a line integral and take a real part, which has seemed difficult given the square root term. 
Any ideas? Appreciate your help!
 A: Fix $x > 1+\tau$ and let $y \in \mathbb{R}$.
Since $x-z\ne 0$ for every $z\in E_\tau$, we have 
\begin{align}
\int_{E_{\tau}} \log |x+iy - z| d\mu_{\tau}(z)&= \int_{E_{\tau}} \log\left|(x-z)\left(1+\frac{iy}{x-z}\right)\right| d\mu_{\tau}(z)\\
&=\int_{E_{\tau}}\log |x-z| d\mu_{\tau}(z)+\int_{E_{\tau}} \log\left|1+\frac{iy}{x-z}\right| d\mu_{\tau}(z).
\end{align}
For small $y$ we have the expansion of $\log\left(1+\frac{iy}{x-z}\right)$ :
$$
\log\left(1+\frac{iy}{x-z}\right)=\frac{iy}{x-z}-\frac{1}{2}\left(\frac{iy}{x-z}\right)^2+O(y^3).
$$
Therefore 
\begin{align}
\int_{E_{\tau}} \log\left|1+\frac{iy}{x-z}\right| d\mu_{\tau}(z)&=\operatorname{Re} \int_{E_{\tau}} \log\left(1+\frac{iy}{x-z}\right) d\mu_{\tau}(z)\\
&=-y\operatorname{Im}\int_{E_{\tau}} \frac{1}{x-z}d\mu_{\tau}(z)+\frac{y^2}{2}\operatorname{Re}\int_{E_{\tau}}\frac{1}{(x-z)^2}d\mu_{\tau}(z)+O(y^3).
\end{align}
Since $$
\int_{E_{\tau}}\left(\operatorname{Im} \frac{1}{x-z}\right)d\mu_{\tau}(z)=0
$$
by the symmetry, we have 
\begin{align}
J(x,y)=J(x,0)+\frac{y^2}{2}\left(A-\frac{1}{1-\tau}\right)+O(y^3),
\end{align}
where $$A=\operatorname{Re}\int_{E_{\tau}}\frac{1}{(x-z)^2}d\mu_{\tau}(z).$$
If we could luckily  prove that $
A-\frac{1}{1-\tau}<0,$ we would have
$$
J(x,y)\le J(x,0)
$$
for sufficiently small $y$.
Addendum:  Evaluation of $A$.   
Let $z=u+iv$. Then
\begin{align}
A&=\operatorname{Re}\frac{1}{\pi (1-\tau^2)}\iint_{E_{\tau}}\frac{1}{(x-u-iv)^2}dudv\\
&=\operatorname{Re}\frac{1}{\pi (1-\tau^2)}\int_{-(1+\tau)}^{1+\tau}du\,\int_{-(1-\tau)\sqrt{1-(u/(1+\tau))^2}}^{(1-\tau)\sqrt{1-(u/(1+\tau))^2}}\frac{1}{(x-u-iv)^2}dv\\
&=\frac{2}{\pi}\int_{-(1+\tau)}^{1+\tau}\frac{\sqrt{(1+\tau)^2-u^2}}{(1+\tau)^2(x-u)^2+(1-\tau)^2((1+\tau)^2- u^2)}du\\
&<\frac{2}{\pi}\int_{-(1+\tau)}^{1+\tau}\frac{\sqrt{(1+\tau)^2-u^2}}{(1+\tau)^2(1+\tau-u)^2+(1-\tau)^2((1+\tau)^2- u^2)}du\\
&=\frac{1}{\pi}\int_{-(1+\tau)}^{1+\tau}\frac{1}{(1+\tau)(1+\tau^2)-2\tau u}\sqrt{\frac{1+\tau+u}{1+\tau-u}}du
\\
&=\frac{1}{\pi}\int_{-1}^{1}\frac{1}{1+\tau^2-2\tau t}\sqrt{\frac{1+t}{1-t}}dt\quad \left(u=(1+\tau)t\right)\\
&=\frac{1}{1-\tau}.
\end{align}
The proof is complete.
