RU-SPB 2009 Problem 117

$f(x)=x^2+x$ $b_1,...,b_{10000}>0$ and $|b_{n+1}-f(b_n)|\leq \frac{1}{1000}$ for $n=1,...,9999$ Prove, that there is such $a_1>0$ that $a_{n+1}=f(a_n);n=1,...,9999$ and $|a_n-b_n|<\frac{1}{100}$