Example 3 Find the maximum value of the function $\gamma=\sqrt{2} \sin \alpha+\sqrt{3} \sin \beta+\sqrt{6} \sin \gamma$, where $\alpha+\beta+\gamma=$ $180^{\circ}$, and find the angles $\alpha, \beta, \gamma$ at which the maximum value is achieved.
In Theorem 2, let $x=\sqrt{2}, y=\sqrt{3}, z=\sqrt{6}$, and substitute into equation (※), and simplify to get $w^{3}+3 w^{2}-3=0$. According to the formula for the roots of a cubic equation, the positive root can be found as $w=\frac{\sqrt{3}}{2} \sec 10^{\circ}$. Substituting $x=\sqrt{2}, y=\sqrt{3}, z=\sqrt{6}, w=\frac{\sqrt{3}}{2} \sec 10^{\circ}$ into inequality $\left(10^{\prime}\right)$, and after simplification, we get
$$\sqrt{2} \sin \alpha+\sqrt{3} \sin \beta+\sqrt{6} \sin \gamma \leqslant \frac{\sqrt{2}}{4}\left(\csc 10^{\circ}\right)^{\frac{3}{2}}$$

Therefore, the function has a maximum value of $\frac{\sqrt{2}}{4}\left(\csc 10^{\circ}\right)^{\frac{3}{2}}$, at which time
$$\cos \alpha=\frac{\sqrt{6}}{4} \sec 10^{\circ}, \cos \beta=\frac{1}{2} \sec 10^{\circ}, \cos \gamma=\frac{\sqrt{2}}{4} \sec 10^{\circ}$$

That is,
$$\alpha=\arccos \left(\frac{\sqrt{6}}{4} \sec 10^{\circ}\right), \beta=\arccos \left(\frac{1}{2} \sec 10^{\circ}\right), \gamma=\arccos \left(\frac{\sqrt{2}}{4} \sec 10^{\circ}\right)$$

Finally, we provide an inequality concerning the relationship between the side lengths and areas of two convex quadrilaterals in a plane.