Suppose $Y$ is a pair of pants with a hyperbolic structure and $\gamma_i; i = 1, 2, 3$ are the geodesic boundaries of length $l_i; i=1, 2, 3$ respectively. Now consider a essential simple arc $\sigma$ in $Y$ with end points on a same boundary component of $Y$ and $l$ denote the length of unique geodesic in the homotopy class of $\sigma$. Then, my question is weather the inequality, $l$ $\geq$ $\frac{1}{2}$ min{$l_i  i= 1, 2, 3$} hold or not for every pair of pants $Y$.
2 Answers
If you keep the lengths of two cuffs fixed (say equal to L) and let the third one go to infinity (say equal to R), then the length $l$ of the orthogeodesic with endpoints on the third cuff goes to zero. There is a trigonometric formula to see this. Cut the pants into two rightangled hexagons then further into four pentagons with the orthogeodesic. Formula 2.6.17 for rightangled pentagons in Chapter 2 of Thurston's notes gives $\sinh(R/4) \sinh (l/2) = \cosh(L/2)$. Therefore no such bound holds.
No it does not. Suppose that the three boundary components have equal and very long length $R$. Then the pair of pants is almost isometric to a graph having two vertices $V,W$ and three edges of length $R/2$ each connecting $V$ to $W$; by "almost isometric" I mean that there is a function from the pants to the graph which distorts length of closed geodesics and of geodesics hitting the boundary at right angles by an additive amount which is bounded for $R$ near $+\infty$. Each of the three boundary curves has its length preserved in this graph. However, the arc you ask about maps to a single edge, of length $R/2$, so back in the pants this arc has length as close to $R/2$ as you like.
This can be made rigorous using formulas of hyperbolic trigonometry found, for example, in Thurston's book The geometry and topology of 3manifolds.

$\begingroup$ I have edited the question. sorry for inconvenience. $\endgroup$ May 24, 2013 at 4:51