A periodic integral inequality (This problem comes in connection with a geometric problem exposed here.)
Let $\gamma(x,y)$ be a (real) function on the unit disk such that
\begin{align}
\frac{\partial^2\gamma}{\partial x \partial y}=0\:\:\:\:\text{and}\:\:\:\:0<\gamma(x,y)<\pi,
\end{align}
then show that

\begin{align}
\int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\sqrt{1-\sin 2t \cos\gamma(t)}\leq 1
\end{align}

being $\gamma(t)=\gamma\left(\cos t, \sin t\right)$ the boundary value of $\gamma(x,y)$.
(For the case of $\gamma(t)$ sufficiently close to $\pi/2$, Robert Bryant's great answer to this question proves the result.)
An attempt of proof:
Following Bryant's idea a possible proof would come as follows. We define the function
\begin{equation}
f(\lambda)\equiv \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\sqrt{1-\sin 2t \sin\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right]},
\end{equation}
being $f(1)$ the desired integral. It follows that
\begin{equation}
f'(0)=-\frac{1}{2} \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\left(\frac{\pi}{2}-\gamma(t)\right)\sin2t=0,
\end{equation}
with the zero in the r.h.s coming only because $\frac{\partial^2\gamma}{\partial x \partial y}=0$. It also follows that $f''(0)<0$. So in order to prove that $f(1)\leq f(0)=1$ we need to show that $f'(\lambda)$ remains negative, i.e
\begin{equation}
f'(\lambda)=-\frac{1}{2} \int_0^{2\pi}\frac{\mathbb{d}t}{2\pi}\frac{\sin2t \cos\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right] }{\sqrt{1-\sin 2t \sin\left[\lambda\left(\frac{\pi}{2}-\gamma(t)\right)\right]}}\cdot\left(\frac{\pi}{2}-\gamma(t)\right)\leq 0 \tag{*} 
\end{equation}
for $\lambda\in (0,1)$. Now, the second derivative is in general not negative in the whole interval, and the integrand in $f'(\lambda)$ oscillates around $0$, so it is apparently highly non-trivial to show $(^*)$
 A: It is false in general. Let $\gamma(x,y)=f(x)+g(y)$. Splitting each term into the odd and even part, we get $\gamma=A(x,y)+X(x)+Y(y)$ where $A$ is even in each variable and $X$ and $Y$ are odd. The condition $0<\gamma<\pi$ implies that
$A(x,y)\pm(|X(x)|+|Y(y)|)\in(0,\pi)$ when $(x,y)\in \mathbb D$. The natural attempt to prove the result would be to use the symmetries of $\sin(2t)$ and reduce the integration to $[0,\pi/2]$ with $x=\cos(t), y=\sin(t)$. Then, expanding $\cos(A+X+Y)$, it would suffice to show that for $x,y>0, x^2+y^2=1$,
$$
\mathcal E_{\varepsilon=(\varepsilon_x,\varepsilon_y)\in\{-1,1\}^2}
\left[1+2xy(\cos A\sin X\sin Y+\varepsilon_x\sin A\cos X\sin Y
\\+\varepsilon_y\sin A\sin X\cos Y-\varepsilon_x\varepsilon_y \cos A\cos X\cos Y)\right]^{1/2}\le 1
$$
for all real numbers $A,X,Y$ satisfying the above assumptions. That would finish the story, but a computer check of random data quickly yields the counterexample
$$
A_0=2,\quad X_0=0.5, 
\\
\quad Y_0=-0.5,\quad              2x_0y_0=0.4
$$
with the average $E=1.01000235275203$ and that is a killer because now we can make $A_1(x)$ and $A_2(y)$ with range in $[\frac\pi 4,1]$ each so that on the quarter-circle, $A(x,y)=A_1(x)+A_2(y)$ is $A_0$ near $(x_0,y_0)$ and quickly descends to $\pi/2$ when you leave that small neighborhood $U$ while $X(x_0)=X_0$, $Y(y_0)=Y_0$ and they quickly drop to $0$ when you leave the neighborhood $U$. In that case you essentially integrate $1$ outside $U$, $E$ on $U$, and the ranges are such that $A(x,y)$ stays in $[\frac\pi 2,2]$ and $|X(x)|, |Y(y)|\le 0.5$ for all $x,y\in[-1,1]$, so the sum $f(x)+g(y)\in[\frac \pi 2-1, 3]\subset (0,\pi)$ in the whole disk and not only on the boundary.
