Rapid convergence is easy to prove, but the
generalization is wrong. The reason seems to be that while $\rho_i <
1$ is true, $\rho_i$ is not bounded away from 1. The line above that
is wrong also, but possibly due to a typo. I offer this
counterexample:
Let $a_0=1$ and let $b_0=4$. Define functions $f$ and $g$ having the
following values for nonnegative integer $i$:
$$f(2-2^{-i},3+2^{-i})=2-2^{-(i+1)};\ g(2-2^{-i},3+2^{-i})=3+2^{-(i+1)}$$
For other values of $x$ and $y$, we only require that the min/max
requirements be satisfied for $f(x,y)$ and $g(x,y)$.
Then $$\lim_{i\to\infinity} a_i =2 \ne \lim_{i\to\infinity} b_i =3$$.
Here is my proof of rapid convergence, which I might have offered
earlier, except that it does not lead to other than a numerical
solution to the second question, ``What is the limit?"
Let $x_i = a_i^2 - b_i^2 = (a_{i-1} + b_{i-1})^2/4 - a_{i-1}b_{i-1} =
(a_{i-1}-b_{i-1})^2/4$. Thus $x_{i+1}/x_i = (1/4) (a_i - b_i)/(a_i +
b_i) < 1/4$. I would start with $b_0=1$ and $a_0$ large, so that $a_i
> b_i$ from the start. Then after $\log a_0$ iterations, convergence
is indeed rapid.