Let $x_1,x_2,\ldots,x_{2001}$ be positive numbers such that $$x_i^2\ge x_1^2+\frac{x_2^2}{2^3}+\frac{x_3^2}{3^3}+\ldots+\frac{x_{i-1}^2}{(i-1)^3}\enspace\text{for }2\le i\le2001.$$Prove that $\sum_{i=2}^{2001}\frac{x_i}{x_1+x_2+\ldots+x_{i-1}}>1.999$.