= Solution
Reverse the order of the two optimizations to obtain the <Lagrangian duality>[Lagrangian dual]
$$
\sup_{y\in\mathbb R^m,\,\gamma\geq0}\ \inf_x\mathcal L(x,y,\gamma).
$$
Put $s=c-A^Ty$ and $r=\sqrt{s^TQ^{-1}s}$. For $\gamma>0$, completing the square gives
$$
x_{\min}=-\frac1{2\gamma}Q^{-1}s,\qquad
q(y,\gamma)=b^Ty-\gamma-\frac{r^2}{4\gamma}.
$$
At $\gamma=0$ the infimum is finite precisely if $s=0$, when it equals $b^Ty$. For $r>0$, the expression $\gamma+r^2/(4\gamma)$ is minimized at $\gamma=r/2$, with value $r$. The same optimized value holds when $r=0$, choosing $\gamma=0$. Eliminating $\gamma$ therefore gives $\sup_y\{b^Ty-\sqrt{s^TQ^{-1}s}\}$. Introducing $\theta=r$ yields
$$
\boxed{\max\{-\theta+b^Ty:A^Ty+s=c,\ s^TQ^{-1}s=\theta^2,\ \theta\geq0\}.}
$$
The nonnegative square root is enforced by $\theta\geq0$. Optimizers correspond by $\gamma=\theta/2$, including the zero case.
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