Let $A_{k+n} = \frac{a_k+a_n}{1+a_k a_n}$. So for any $n$, $A_n = \frac{a_1+a_{n-1}}{1+a_1 a_{n-1}}$. Consider the function
$$f(x) = \frac{a_1 + x}{1 + a_1 x}$$
where $x > 0$. Since $f(x) = \frac{1}{a_1} \frac{a_1^2+a_1x}{1+a_1x} = \frac{a_1(a_1+a_1x)}{a_1+a_1^2x}$, it follows that
$$f(x) \geq \begin{cases} 1/a_1 & \text{if } a_1 \geq 1 \\ a_1 & \text{if } 0 < a_1 < 1. \end{cases}$$
So for any value of $a_1$, there exists $0 < t \le 1$ so that $f(x) \ge t$. (In fact just take $t$ to be the smaller of $a_1$ and $1/a_1$.)
Thus for any $n$, $A(n) \ge t$ and so
$$A_{2n} = A_{n+n} = \frac{2a_n}{1+a_n^2} \ge t \\ \therefore ta_n^2 - 2a_n + t \le 0$$
Thus $a_n$ lies between the 2 roots of the equation $tx^2 - 2x + t = 0$. The roots are $\frac{1\pm\sqrt{1-t^2}}{t}$. Letting $b = \frac{1-\sqrt{1-t^2}}{t}$ and $c = \frac{1+\sqrt{1-t^2}}{t}$ and we are done.