Suppose $A_1,\ldots,A_n$ is a sequence of $2 \times 2$ complex matrices such that $| \det(A_j) | =1$ and $ | \mathrm{tr}(A_j) | > 2 $ for each $j$. What kinds of reasonable restrictions can one place on the $A_j$'s to guarantee that $| \mathrm{tr}(A_n \cdots A_1)| > 2$? Obviously, some restrictions are needed -- for example, the inverse of any hyperbolic $\mathrm{SL}_2(\mathbb R)$ matrix is also hyperbolic, but their product, the identity, is not.

To get some idea for the source of this question, let me point out that one "reasonable restriction" which produces the desired conclusion is that the $A$'s be drawn from the family $$ \mathcal{S} = \left\{ B_t =\begin{pmatrix} t & -1 \\ 1 & 0 \end{pmatrix} : t \in \mathbb R \right\} $$ That is, if $A_j \in \mathcal{S}$ and $|\mathrm{tr}(A_j)| > 2$ for each $j$, then $ |\mathrm{tr}(A_n \cdots A_1) | >2 $. The class $\mathcal{S}$ arises naturally in the study of discrete one-dimensional Hamiltonians $H_V$ which act on $\ell^2(\mathbb{Z})$ via $$ (H_V u)_n = u_{n-1} + u_{n+1} + V_n u_n, $$ where $V$ is some bounded sequence.

Using these operators, I can sketch the argument which proves the claim about $\mathcal{S}$. Suppose $V_j$ is an $n$-periodic sequence. Since $H_V = H_0 + V$, $\|H_0\| = 2$, and $\sigma(V) = \{V_1,\ldots,V_n\}$, we have the following easy inclusion for the spectrum of $H_V$: $$ \sigma(H_V) \subseteq \bigcup_{j=1}^n ([-2,2] + V_j). $$ On the other hand, by a version of Floquet theory for such operators, $$ \sigma(H_V) = \{ E \in \mathbb{R} : | \mathrm{tr}(B_{E-V_n} \cdots B_{E - V_1} )| \leq 2 \} $$ From the inclusion, we see that $|V_j| > 2$ for all $j$ implies $0 \notin \sigma(H_V)$. Using the equality from Floquet theory, we get $ |\mathrm{tr}(B_{-V_n} \cdots B_{-V_1})| > 2 $.

EDIT: Of course, I'm sure that one probably doesn't need the operator-theoretic argument to prove that the product of hyperbolic elements of $\mathcal{S}$ is hyperbolic -- one can probably prove this statement by a direct inductive argument.