设 $a_i,b_i>0$（$1\leqslant i\leqslant n+1$），$b_{i+1}-b_i\geqslant \delta >0$（$i=1,2,\cdots,n$，$\delta$ 为常数）．若 $\displaystyle\sum_{i=1}^n a_i=1$，证明：\[\sum_{i=1}^n\dfrac{\sqrt[i]{a_1a_2\cdots a_ib_1b_2\cdots b_i}}{b_{i+1}b_i}<\dfrac1{\delta}.\]
解析 记 $s_k=\displaystyle\sum_{i=1}^ka_ib_i$，$s_0=0$，则有\[a_k=\dfrac{s_k-s_{k-1}}{b_k},\quad k=1,2,\cdots,n+1,\]因此有\[\begin{split} 1&=\sum_{i=1}^na_i=\sum_{i=1}^n\dfrac{s_i-s_{i-1}}{b_i}\\ &=\sum_{i=1}^n\left(\dfrac{s_i}{b_i}-\dfrac{s_i}{b_{i+1}}\right)+\dfrac{s_{n+1}}{b_{n+1}}\\ &\geqslant \delta\sum_{i=1}^n\dfrac{s_i}{b_ib_{i+1}}\\ &\geqslant \delta \sum_{i=1}^n\dfrac{i\sqrt[i]{a_1a_2\cdots a_ib_1b_2\cdot b_i}}{b_ib_{i+1}},\end{split}\]其中依次用到了阿贝尔求和以及均值不等式，原命题得证．