(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

$$ \frac{\partial^2\gamma}{\partial x \, \partial y} = 0\:\:\:\:\text{and}\:\:\:\:0<\gamma(x,y)<\pi, $$

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 $(*)$