已知 $\triangle ABC$ 的三个内角分别为 $A,B,C$ 且 $C\geqslant \dfrac{\pi}3$，求证：$\Big(a+b\Big)\Big(\dfrac 1a+\dfrac 1b+\dfrac 1c\Big)\geqslant 4+\dfrac{1}{\sin\dfrac C2}$．
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【解析】
不妨设 $A\geqslant B$，令 $A=x+y$，$B=x-y$，则\[0\leqslant y<x\leqslant \dfrac{\pi}3.\]设\[M=\Big(a+b\Big)\Big(\dfrac 1a+\dfrac 1b+\dfrac 1c\Big)- 4-\dfrac{1}{\sin\dfrac C2},\]则\[\begin{split} M&=\dfrac ab+\dfrac ba+\dfrac{a+b}c-2-\dfrac{1}{\sin\dfrac C2}\\
&=\dfrac{\sin(x+y)}{\sin(x-y)}+\dfrac{\sin(x-y)}{\sin(x+y)}+\dfrac{\sin(x+y)+\sin(x-y)}{\sin 2x}-2-\dfrac{1}{\cos x}\\
&=\dfrac{\sin^2(x+y)+\sin^2(x-y)-2\sin(x-y)\sin(x+y)}{\sin(x-y)\sin(x+y)}+\dfrac{2\sin x\cos y}{2\sin x\cos x}-\dfrac{1}{\cos x}\\
&=\dfrac{2\sin^2x\cos^2y+2\cos^2x\sin^2y-2\sin^2x+2\sin^2y}{\sin^2x-\sin^2y}-\dfrac{1-\cos y}{\cos x}\\
&=\dfrac{4\cos^2x\sin^2y}{\sin^2x-\sin^2y}-\dfrac{1-\cos y}{\cos x}\\
&=\dfrac{2\sin^2\dfrac y2}{\cos x\left(\sin^2x-\sin^2y\right)}\cdot \left(8\cos^2\dfrac y2\cos^3x-\sin^2x+\sin^2y\right)\\
&\geqslant \dfrac{2\sin^2\dfrac y2}{\cos x\left(\sin^2x-\sin^2y\right)}\cdot\left(8\cos^2\dfrac{\pi}6\cos^2\dfrac{\pi}3-\sin^2\dfrac{\pi}3\right)\\
&\geqslant 0,\end{split}\]因此原命题得证．
&=\dfrac{\sin(x+y)}{\sin(x-y)}+\dfrac{\sin(x-y)}{\sin(x+y)}+\dfrac{\sin(x+y)+\sin(x-y)}{\sin 2x}-2-\dfrac{1}{\cos x}\\
&=\dfrac{\sin^2(x+y)+\sin^2(x-y)-2\sin(x-y)\sin(x+y)}{\sin(x-y)\sin(x+y)}+\dfrac{2\sin x\cos y}{2\sin x\cos x}-\dfrac{1}{\cos x}\\
&=\dfrac{2\sin^2x\cos^2y+2\cos^2x\sin^2y-2\sin^2x+2\sin^2y}{\sin^2x-\sin^2y}-\dfrac{1-\cos y}{\cos x}\\
&=\dfrac{4\cos^2x\sin^2y}{\sin^2x-\sin^2y}-\dfrac{1-\cos y}{\cos x}\\
&=\dfrac{2\sin^2\dfrac y2}{\cos x\left(\sin^2x-\sin^2y\right)}\cdot \left(8\cos^2\dfrac y2\cos^3x-\sin^2x+\sin^2y\right)\\
&\geqslant \dfrac{2\sin^2\dfrac y2}{\cos x\left(\sin^2x-\sin^2y\right)}\cdot\left(8\cos^2\dfrac{\pi}6\cos^2\dfrac{\pi}3-\sin^2\dfrac{\pi}3\right)\\
&\geqslant 0,\end{split}\]因此原命题得证．
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