Are there some weaker than concavity conditions on a function $\gamma(x,y)$ satisfying $0 < \gamma(x,y) \leq 1$ with $\gamma(x,y) = 1 \iff x = y$ such that the function $$ f(x,y) = \begin{pmatrix} x \\ y \end{pmatrix}^\top \begin{pmatrix} 1 & \gamma(x, y) \\ \gamma(x, y) & 1 \end{pmatrix}^{-1} \begin{pmatrix} x \\ y \end{pmatrix} $$ is quasiconvex?
P.S. If $\gamma$ is concave, then since $g(\boldsymbol{x}, A) = \boldsymbol{x}^\top A^{-1} \boldsymbol{x}$ is convex and monotone in both arguments, the convexity of $f$ follows from the composition theorem. I'm interested in whether there are other meaningful conditions on $\gamma$, which guarantee this. I recently asked this question with $\sin$ function, but quickly realised that I can prove that it's not in fact quasiconvex, so I deleted the question.
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