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已知函数 ${\displaystyle f(x)=(1+x{)}^m-mx-1,x\in (-1,+\mathrm{\infty }),m>0}$ 且 ${\displaystyle m\ne 1}$.

1. 讨论 ${\displaystyle f(x)}$ 的单调性;
2. 若 ${\displaystyle {\displaystyle \mathrm{\forall }x\in (0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )}}$, ${\displaystyle a\sin x<(1+{\cos }^{2}x{)}^{\sin x}}$, 求 ${\displaystyle a}$ 的取值范围;
3. 证明: 当 ${\displaystyle {\displaystyle x\in (0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )}}$, 且 ${\displaystyle n\in \mathbb{N},n\ge 2}$ 时，恒有 $${\displaystyle \frac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots (1-\sqrt[n]{\sin x})}{(1-\sin x{)}^{n-1}}>\frac{1}{n!}.}$$

1. 解 ${\displaystyle f^{\prime }(x)=m[(1+x{)}^{m-1}-1]}$. ${\displaystyle 0<m<1}$ 时 ${\displaystyle f(x)}$ 在 ${\displaystyle (-1,0)}$ 上减，在 ${\displaystyle (0,+\mathrm{\infty })}$ 上增；${\displaystyle m>1}$ 时 ${\displaystyle f(x)}$ 在 ${\displaystyle (-1,0)}$ 上增，在 ${\displaystyle (0,+\mathrm{\infty })}$ 上减.

2. 解 当 ${\displaystyle a\le 0}$，不等式显然成立；当 ${\displaystyle a>0}$，记 ${\displaystyle t=\sin x\in (0,1)}$，则原不等式等价于$$\begin{array}{r}at<(2-t^2{)}^t\Rightarrow \ln a<t\ln (2-t^2)-\ln t,\end{array}$$
   记右侧函数为${\displaystyle g(t),t\in (0,1)}$，则$$\begin{array}{rl}& g^{\prime }(t)=\ln (2-t^2)-\frac{2t^2}{2-t^2}-\frac{1}{t}\\ \le & (2-t^2)-1-\frac{2t^2}{2-t^2}-\frac{1}{t}=\frac{-t^2(4-t^2)}{2-t^2}+\frac{t-1}{t}<0,\end{array}$$
   故只需${\displaystyle \ln a\le g(1)=0\Rightarrow a\le 1}$.

   综上，${\displaystyle a\le 1}$.

3. 证明 记 ${\displaystyle \sqrt[n]{\sin x}=t<1}$，则$$\begin{array}{r}\frac{1-\sqrt[n]{\sin x}}{1-\sin x}=\frac{1-t}{1-t^n}=\frac{1}{1+t+\cdots +t^{n-1}}>\frac{1}{n},\end{array}$$
   累乘即得$$\begin{array}{r}\prod _{i=2}^n\frac{1-\sqrt[i]{\sin x}}{1-\sin x}=\frac{\prod _{i=2}^n(1-\sqrt[i]{1-\sin x})}{(1-\sin x{)}^{n-1}}>\frac{1}{n!}.\end{array}$$