25-26高三上·山东潍坊期中·第14题

$3\sqrt{3}$

【题目】已知锐角$\alpha, \beta, \gamma$满足$\alpha + \beta + \gamma = \frac{\pi}{2}$，若不等式$\sin\alpha\sin\beta(k\sin\gamma – \cos\gamma)$$\leq \sin\alpha\cos\beta\sin\gamma + \cos\alpha\sin\beta\sin\gamma$恒成立，则实数$k$的最大值为________．

【答案】$3\sqrt{3}$

【解析】$\because 0 < \alpha, \beta, \gamma < \frac{\pi}{2}$，$\therefore \sin\alpha\sin\beta\sin\gamma > 0$，
将左边括号展开移项后两边同除$\sin\alpha\sin\beta\sin\gamma$可得$k \leq \frac{1}{\tan\alpha} + \frac{1}{\tan\beta} + \frac{1}{\tan\gamma} $$= \frac{\sin(\alpha + \beta)}{\sin\alpha\sin\beta} + \frac{\cos\gamma}{\sin\gamma}$，

化简得右式$= \frac{\sin\left(\frac{\pi}{2} – \gamma\right)}{\sin\alpha\sin\beta} + \frac{\cos\gamma}{\sin\gamma} $$= \frac{\cos\gamma}{\sin\alpha\sin\beta} + \frac{\cos\gamma}{\sin\gamma}$
$= \frac{2\cos\gamma}{\cos(\alpha – \beta) – \cos(\alpha + \beta)} + \frac{\cos\gamma}{\sin\gamma} $$= \frac{2\cos\gamma}{\cos(\alpha – \beta) – \cos\left(\frac{1}{2}\pi – \gamma\right)} + \frac{\cos\gamma}{\sin\gamma}$
$= \frac{2\cos\gamma}{\cos(\alpha – \beta) – \sin\gamma} + \frac{\cos\gamma}{\sin\gamma}$，

以$\alpha – \beta$作为主元可得右式在$\alpha = \beta$时取到最小值．

此时右式$= \frac{\sin(\alpha + \beta)}{\sin\alpha\sin\beta} + \frac{\cos\gamma}{\sin\gamma}$$= \frac{\sin2\alpha}{\sin\alpha\sin\alpha} + \frac{\cos\left(\frac{\pi}{2} – 2\alpha\right)}{\sin\left(\frac{\pi}{2} – 2\alpha\right)}$$= \frac{2\sin\alpha\cos\alpha}{\sin\alpha\sin\alpha} + \frac{\sin2\alpha}{\cos2\alpha}$$= \frac{2}{\tan\alpha} + \tan2\alpha$．

令$\tan\alpha = t$，则右边$= \frac{2}{t} + \frac{2t}{1 – t^2} = \frac{2}{t – t^3}$．

令$f(t) = t – t^3, t > 0$，$f'(t) = 1 – 3t^2, t > 0$，
即函数$f(t)$在$\left(0, \frac{\sqrt{3}}{3}\right)$上单调递增，在$\left(\frac{\sqrt{3}}{3}, +\infty\right)$上单调递减，
$\therefore f(t)_{\max} = f\left(\frac{\sqrt{3}}{3}\right) = \frac{2}{3\sqrt{3}}$，因此$k \leq 3\sqrt{3}$．

故答案为：$3\sqrt{3}$．