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\[
\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{2a^2+3c^2-2\sqrt{2}ac}{2ac}\geqslant \dfrac{\sqrt{6}-\sqrt{2}}{4}.\]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\(2\sqrt3+2\).
原文：https://www.cnblogs.com/Math521/p/11959797.html