看着好像很宽松的不等式，证明是不是很粗糙啊？

郝酒

$a_i,b_i\in R^+$求证
$$\sum_{k=1}^{n}\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^{n}\frac{b_i^2}{a_i}$$
很早以前抄的题目，到现在也不会解，求助一下大侠们。
另外如果有人知道这题的出处，就再好不过啦！

郝酒

令$b_i=1$,可以得到一个简单的不等式，用放缩的方法可以做。
$$\sum\frac{k}{a_1+\cdots+a_k}<2\sum\frac{1}{a_i}$$

郝酒

再顶一番，ku版出马救一下咯：）

郝酒

这有一个数学归纳法的解法，挺粗暴的：）artofproblemsolving.com/community/u113828h1289401p6815724
数归证这个加强不等式：
$$\begin{align*}
\sum_{k=1}^{n} & \frac{\left(b_{1}+b_{2}+\cdots+b_{k}\right)b_{k}}{a_{1}+a_{2}+\cdots+a_{k}}\leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\sum_{k=2}^{n}\frac{b_{k}^{2}}{a_{k}}.
\end{align*}$$
$n=1$时显然,假设对$n-1$也成立,则有
$$\begin{align*}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & +\sum_{k=3}^{n}\frac{\left(b_{1}+\cdots+b_{k}\right)b_{k}}{a_{1}+\cdots+a_{k}}\leq\frac{3}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}}+2\sum_{k=3}^{n}\frac{b_{k}^{2}}{a_{k}}
\end{align*}$$
$$\begin{align*}
\frac{b_{1}^{2}}{a_{1}}+\frac{\left(b_{1}+b_{2}\right)b_{2}}{a_{1}+a_{2}}+\frac{1}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & \leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\frac{b_{2}^{2}}{a_{2}}\\
\frac{b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2}}{a_{1}+a_{2}} & \leq\frac{b_{1}^{2}}{a_{1}}+4\frac{b_{2}^{2}}{a_{2}}\\
b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2} & \leq b_{1}^{2}+\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+4b_{2}^{2}\\
4b_{1}b_{2} & \leq\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+b_{2}^{2}
\end{align*}$$
由
$$\begin{align*}
\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2} & \geq2\sqrt{\frac{a_{2}}{a_{1}}b_{1}^{2}\cdot4\frac{a_{1}}{a_{2}}b_{2}^{2}}\\
& =4b_{1}b_{2}
\end{align*}$$
即得.