设 $a_i,b_i>0$（$1\leqslant i\leqslant n+1$），$b_{i+1}-b_t\geqslant \delta >0$（$\delta$ 为常数）．若 $\displaystyle \sum^n_{k=1}a_i=1$，证明：$$\sum^n_{i=1}\frac{i\sqrt[i]{a_1a_2\ldots a_ib_1b_2\ldots b_i}}{b_{i+1}b_i}<\frac{1}{\delta}.$$
【难度】
【出处】
无
【标注】
【答案】
略
【解析】
记 $\displaystyle s_k=\sum^k_{i=1}a_ib_i, s_0=0$，则有 $a_k=\frac{s_k-s_{k-1}}{b_k}$．
由已知$$\begin{aligned}
1&=\sum^k_{i=1}a_i=\sum^n_{i=1}\frac{s_i-s_{i-1}}{b_i}\\
&=\sum^n_{i=1}(\frac{s_i}{b_i}-\frac{s_i}{b_{i+1}})+\frac{s_{n+1}}{b_{n+1}}\geqslant \delta \sum^n_{i=1}\frac{s_i}{b_ib_{i+1}}\\
&\geqslant \delta \sum^n_{i=1}\frac{i\sqrt[i]{A_1A_2\ldots a_ib_1b_2\ldots b_i}}{b_ib_{i+1}} (\text{因为}is_i\geqslant \sqrt[i]{A_1A_2\ldots a_ib_1b_2\ldots b_i}),\\
\end{aligned}$$即$$\sum^n_{i=1}\frac{i\sqrt[i]{a_1a_2\ldots a_ib_1b_2\ldots b_i}}{b_{i+1}b_i}<\frac{1}{\delta}.$$
由已知$$\begin{aligned}
1&=\sum^k_{i=1}a_i=\sum^n_{i=1}\frac{s_i-s_{i-1}}{b_i}\\
&=\sum^n_{i=1}(\frac{s_i}{b_i}-\frac{s_i}{b_{i+1}})+\frac{s_{n+1}}{b_{n+1}}\geqslant \delta \sum^n_{i=1}\frac{s_i}{b_ib_{i+1}}\\
&\geqslant \delta \sum^n_{i=1}\frac{i\sqrt[i]{A_1A_2\ldots a_ib_1b_2\ldots b_i}}{b_ib_{i+1}} (\text{因为}is_i\geqslant \sqrt[i]{A_1A_2\ldots a_ib_1b_2\ldots b_i}),\\
\end{aligned}$$即$$\sum^n_{i=1}\frac{i\sqrt[i]{a_1a_2\ldots a_ib_1b_2\ldots b_i}}{b_{i+1}b_i}<\frac{1}{\delta}.$$
答案
解析
备注
