回復 31# leo790124 的帖子
\(\begin{align}
& \cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{{{a}^{2}}+{{b}^{2}}-\frac{{{a}^{2}}+{{b}^{2}}}{3}}{2ab}=\frac{1}{3}\times \frac{{{a}^{2}}+{{b}^{2}}}{ab}\ge \frac{2}{3} \\
& \sin C=\sqrt{1-{{\cos }^{2}}C}=\frac{\sqrt{5}}{3} \\
\end{align}\)