已知( riangle 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{qquadqquad}.)
解析:
由题有$$
2b=sqrt{2}a+c.$$因此$$
cos B=dfrac{a2+c2-b2}{2ac}=dfrac{2a2+3c^2-2sqrt{2}ac}{2ac}geqslant dfrac{sqrt{6}-sqrt{2}}{4}.$$因此$$sin B=sqrt{1-cos^2B}leqslant dfrac{sqrt{6}+sqrt{2}}{4}.$$
由正弦定理我们有$$
sqrt2 sin A+sin C=2sin Bleqslant dfrac{sqrt{6}+sqrt{2}}{2}.$$记待求表达式为(M),则
[ egin{split}
M&=sqrt{2}cdotleft(dfrac{3}{sqrt{2}sin A}+dfrac{1}{sin C}
ight)\
&geqslant sqrt{2}cdot dfrac{left(sqrt{3}+1
ight)^2}{sqrt{2}sin A+sin C}\
&geqslant 2left(sqrt{3}+1
ight).
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$2sqrt3+2$.