HMMT November, 2018

Solution:
Answer: $\frac{4}{3}$
Let $a_{0}=a$ and $b_{0}=b$. From $\sum_{i=0}^{\infty} a_{i}=\frac{a_{0}}{1-r_{a}}=1$ we have $a_{0}=1-r_{a}$ and similarly $b_{0}=1-r_{b}$. This means $\sum_{i=0}^{\infty} a_{i}^{2}=\frac{a_{0}^{2}}{1-r_{a}^{2}}=\frac{a^{2}}{\left(1-r_{a}\right)\left(1+r_{a}\right)}=\frac{a^{2}}{a(2-a)}=\frac{a}{2-a}$, so $\sum_{i=0}^{\infty} a_{i}^{2} \sum_{i=0}^{\infty} b_{i}^{2}=\sum_{i=0}^{\infty} a_{i} b_{i}$ yields
$$\frac{a}{2-a} \cdot \frac{b}{2-b}=\frac{a b}{1-(1-a)(1-b)} .$$
Since the numerators are equal, the denominators must be equal, which when expanded gives $2 a b-3 a-3 b+4=0$, which is equivalent to $(2 a-3)(2 b-3)=1$. But note that $0<a, b<2$ since we need the sequences to converge (or $\left|r_{a}\right|,\left|r_{b}\right|<1$), so then $-3<2 b-3<1$, and thus $2 a-3>1$ (impossible) or $2 a-3<-\frac{1}{3}$. Hence $a<\frac{4}{3}$, with equality when $b$ approaches $0$.