Length of geodesic representative on hyperbolic surfaces Let $S$ be a closed oriented hyperbolic surface. Let $x,y \in S$ and let $\alpha,\beta$ be two geodesic arcs with endpoints $x$ and $y$. Let $\alpha \beta$ be the closed piecewise geodesic curve obtained from the concatenation of $\alpha$ and $\beta$ and let $\gamma$ be the unique closed geodesic freely homotopic to $\alpha \beta$.
Question: is it possible to bound the length of $\gamma$ from below in terms of the length of $\alpha$ and $\beta$?
 A: There is no bound of this sort. Or, to put it more precisely, the only bound you can have of this sort is the length of a shortest simple closed geodesic, which is a constant for the surface and which does not depend in any way on the lengths of $\alpha$ and $\beta$.
To see why, start from $\gamma$ being a shortest simple closed geodesic on $S$, which implies that $\gamma$ is a shortest closed geodesic (whether simple or not). 
Let $x,z \in \gamma$ be antipodal points of $\gamma$, meaning that they subdivide $\gamma$ into two equal segments. Denote those two segments $\gamma_1,\gamma_2$ with initial endpoint $x$ and terminal endpoint $z$. Let $\delta$ be a geodesic path with initial endpoint $z$ and terminal endpoint $y$; this path $\delta$ can be as long as you like, a gajillion bazillion light years long. 
Now let $\alpha$ be the geodesic path which is path homotopic to $\gamma_1 \delta$ and let $\beta$ be the geodesic path which is path homotopic to $\gamma_2 \delta$. Each of these two paths has length greater than or equal to $\text{Length}(\delta) - \frac{1}{2} \text{Length}(\gamma)$, which can be as long as you like.
Now concatenate $\alpha$ and $\beta$ (you wrote the concatenation incorrectly as $\alpha\beta$, so let me correct that to $\alpha\bar\beta$). The conclusion is that the unique closed geodesic that is freely homotopic to $\alpha\bar\beta$ is simply $\gamma$, which is the shortest closed geodesic of them all.
