一道特殊的三角最值问题

题目

在 \(\triangle ABC\) 中，\(\sqrt{2}\sin A+\sin B\sin C\) 的最大值是 \((\qquad)\)

\(A. 2+\dfrac{1}{2}\)

\(B. 2\)

\(C. \sqrt{3}\)

\(D. \sqrt{5}\)

解析

\[\begin{array}{rl}\sqrt{2}\sin A+\sin B\sin C&=\sqrt{2}\sin A-\dfrac{\cos(B+C)-\cos(B-C)}{2}\\&=\sqrt{2}\sin A+\dfrac{\cos(B-C)-\cos(B+C)}{2}\\&\leq\sqrt{2}\sin A+\dfrac{1+\cos A}{2}\\&=\sqrt{2}\sin A+\dfrac{1}{2}\cos A+\dfrac{1}{2}\\&\leq\sqrt{2+\dfrac{1}{4}}+\dfrac{1}{2}=2\end{array} \]

当且仅当 \(\sin B=\sin C=\dfrac{\sqrt{6}}{3},\sin A=\dfrac{2\sqrt{2}}{3}\) 时，等号成立.

答案 \(B\)