设 $x\geqslant y\geqslant z\geqslant \dfrac{\pi}{12}$，且 $x+y+z=\dfrac{\pi}2$，求乘积 $\cos x\cos y\cos z$ 的最大值与最小值．
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【答案】
最大值为 $\dfrac{2+\sqrt3}{8}$，最小值为 $\dfrac18$
【解析】
记所求表达式为 $M$，先求最大值
由于 $x\geqslant y\geqslant z\geqslant \dfrac{\pi}{12}$，$x+y+z=\dfrac{\pi}2$，所以$$\begin{split} M&\leqslant \cos y\cos z\cos\dfrac{\pi}{12}\\
&=\dfrac12\sin2y\cos\dfrac{\pi}{12}\\
&\leqslant \dfrac12\sin(x+y)\cos\dfrac{\pi}{12}\\
&\leqslant \dfrac12\sin\left(\dfrac{\pi}2-\dfrac{\pi}{12}\right)\cos\dfrac{\pi}{12}\\
&=\dfrac{2+\sqrt3}{8}. \end{split}$$故当 $(x,y,z)=\left(\dfrac{5\pi}{24},\dfrac{5\pi}{24},\dfrac{\pi}{24}\right)$ 时，所求表达式取得最大值 $\dfrac{2+\sqrt3}{8}$．
再求最小值
由于 $x\geqslant y\geqslant z\geqslant \dfrac{\pi}{12}$，$x+y+z=\dfrac{\pi}{2}$，所以$$\left(\dfrac{\pi}{12}\leqslant z\leqslant y\leqslant x\leqslant \dfrac{\pi}{3}\right)\land\left(0\leqslant x-z\leqslant \dfrac{\pi}{4}\right).$$于是$$\begin{split}M &=\dfrac12[\cos(x+z)+\cos(x-z)]\sin y\\
&\geqslant \dfrac12[\sin y+\cos\dfrac{\pi}{4}]\sin y\\
&=\dfrac12\left(\sin y+\dfrac12\cos\dfrac{\pi}{4}\right)^2-\dfrac18\cos^2\dfrac{\pi}{4}\\
&\geqslant \dfrac12\left(\sin\dfrac{\pi}{12}+\dfrac12\cos\dfrac{\pi}4\right)^2-\dfrac18\cos^2\dfrac{\pi}{4}\\
&=\dfrac18. \end{split}$$当 $(x,y,z)=\left(\dfrac{\pi}3,\dfrac{\pi}{12},\dfrac{\pi}{12}\right)$ 时，所求表达式取得最小值 $\dfrac18$．
由于 $x\geqslant y\geqslant z\geqslant \dfrac{\pi}{12}$，$x+y+z=\dfrac{\pi}2$，所以$$\begin{split} M&\leqslant \cos y\cos z\cos\dfrac{\pi}{12}\\
&=\dfrac12\sin2y\cos\dfrac{\pi}{12}\\
&\leqslant \dfrac12\sin(x+y)\cos\dfrac{\pi}{12}\\
&\leqslant \dfrac12\sin\left(\dfrac{\pi}2-\dfrac{\pi}{12}\right)\cos\dfrac{\pi}{12}\\
&=\dfrac{2+\sqrt3}{8}. \end{split}$$故当 $(x,y,z)=\left(\dfrac{5\pi}{24},\dfrac{5\pi}{24},\dfrac{\pi}{24}\right)$ 时，所求表达式取得最大值 $\dfrac{2+\sqrt3}{8}$．
再求最小值
由于 $x\geqslant y\geqslant z\geqslant \dfrac{\pi}{12}$，$x+y+z=\dfrac{\pi}{2}$，所以$$\left(\dfrac{\pi}{12}\leqslant z\leqslant y\leqslant x\leqslant \dfrac{\pi}{3}\right)\land\left(0\leqslant x-z\leqslant \dfrac{\pi}{4}\right).$$于是$$\begin{split}M &=\dfrac12[\cos(x+z)+\cos(x-z)]\sin y\\
&\geqslant \dfrac12[\sin y+\cos\dfrac{\pi}{4}]\sin y\\
&=\dfrac12\left(\sin y+\dfrac12\cos\dfrac{\pi}{4}\right)^2-\dfrac18\cos^2\dfrac{\pi}{4}\\
&\geqslant \dfrac12\left(\sin\dfrac{\pi}{12}+\dfrac12\cos\dfrac{\pi}4\right)^2-\dfrac18\cos^2\dfrac{\pi}{4}\\
&=\dfrac18. \end{split}$$当 $(x,y,z)=\left(\dfrac{\pi}3,\dfrac{\pi}{12},\dfrac{\pi}{12}\right)$ 时，所求表达式取得最小值 $\dfrac18$．
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