设正实数 $a_1,a_2,\cdots,a_{100}$ 满足 $a_i\geqslant a_{101-i}$（$i=1,2,\cdots,50$）．记 $x_k=\dfrac{ka_{k+1}}{a_1+a_2+\cdots+a_k}$（$k=1,2,\cdots,99$）．证明：$x_1x_2^2\cdots x_{99}^{99}\leqslant 1$．
解析 根据题意，有\[\begin{split} LHS&=\prod_{k=1}^{99}\left(\dfrac{k a_{k+1}}{a_1+a_2+\cdots+a_k}\right)^k\\ &=\prod_{k=1}^{99}\left(\left(\dfrac k{a_1+a_2+\cdots+a_k}\right)^k\cdot a_{k+1}^k\right)\\ &\leqslant \prod_{k=1}^{99}\dfrac{a_{k+1}^k}{a_1a_2\cdots a_k}\\ &=\left(\dfrac{a_{100}}{a_1}\right)^{99}\cdot\left(\dfrac{a_{99}}{a_2}\right)^{97}\cdots \left(\dfrac{a_{51}}{a_{50}}\right)^1\\ &\leqslant 1,\end{split}\]命题得证．