已知\(\triangle ABC\)中,\(a,b,c\)分别为三角形三个内角\(A,B,C\)所对的边,\(\sqrt2 a,b,c\)成等差数列,则\(\dfrac{3}{\sin A}+\dfrac{\sqrt{2}}{\sin C}\)的最小值为\(\underline{\qquad\qquad}.\)
解析:
由题有$$
2b=\sqrt{2}a+c.$$因此$$
\cos B=\dfrac{a2+c2-b2}{2ac}=\dfrac{2a2+3c^2-2\sqrt{2}ac}{2ac}\geqslant \dfrac{\sqrt{6}-\sqrt{2}}{4}.$$因此$$\sin B=\sqrt{1-\cos^2B}\leqslant \dfrac{\sqrt{6}+\sqrt{2}}{4}.$$
由正弦定理我们有$$
\sqrt2 \sin A+\sin C=2\sin B\leqslant \dfrac{\sqrt{6}+\sqrt{2}}{2}.$$记待求表达式为\(M\),则
\[ \begin{split}
M&=\sqrt{2}\cdot\left(\dfrac{3}{\sqrt{2}\sin A}+\dfrac{1}{\sin C}\right)\\
&\geqslant \sqrt{2}\cdot \dfrac{\left(\sqrt{3}+1\right)^2}{\sqrt{2}\sin A+\sin C}\\
&\geqslant 2\left(\sqrt{3}+1\right).
\end{split}
\]
以上各处不等式的取等条件一致,均为$\sqrt{2}a=\sqrt{3}c$.因此$M$的最小值为$2\sqrt3+2$.
