Let ($b_n$) be a sequence such that $b_n \geq 0 $ and $b_{n+1}^2 \geq \frac{b_1^2}{1^3}+\cdots+\frac{b_n^2}{n^3}$ for all $n \geq 1$. Prove that there exists a natural number $K$ such that
\[\sum_{n=1}^{K} \frac{b_{n+1}}{b_1+b_2+ \cdots + b_n} \geq \frac{1993}{1000}\]
I don't know if this is right.
\[b_{n+1}^2\ge \frac{(b_1+b_2+\cdots+b_n)^2}{\left(\frac{n(n+1)}{2}\right)^2}\]
\[\Longrightarrow \frac{b_{n+1}}{b_1+\cdots+b_n}\ge\frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}\]
\[\Longrightarrow \[ \sum_{n=1}^{K}\frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}}\geq 2-\frac{2}{k+1} \]
That will be greater than $\frac{1993}{1000}$ for $k\ge 285$.