已知数列 $\left\{x_{n}\right\}$（$n \in \mathbb{N}^{\ast}$）满足 $x_{n}>1$，$\mathrm{e}$ 为自然对数的底数，记\[ \begin{cases} A=x_{1} x_{2} \cdots x_{n}, \\ B=x_{1}+x_{2}+\cdots+x_{n},\\ C=\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}}+\cdots+\dfrac{1}{x_{n}} ,\end{cases}\]证明：$A^{2}<\mathrm{e}^{B-C}$．
解析 根据题意，有\[\begin{split} A^2<{\rm e}^{B-C}&\iff 2\ln A<B-C\\ &\iff 2\ln (x_1x_2\cdots x_n)<(x_1+x_2\cdots+x_n)-\left(\dfrac1{x_1}+\dfrac1{x_2}+\cdots+\dfrac1{x_n}\right)\\ &\iff \sum_{k=1}^n\ln x_k<\dfrac 12\left(x_k-\dfrac{1}{x_k}\right),\end{split}\]根据对数函数的进阶放缩，有\[\ln x<\dfrac 12\left(x-\dfrac 1x\right),\quad x>1,\]因此原命题得证．