Let $c_1:[0,t_1]\rightarrow \mathbb{H}^2$ and $c_2:[0,t_2]\rightarrow \mathbb{H}^2$ be two unit-speed geodesics in the hyperbolic plane with $c_1(0)=c_2(0)$ and such that the angle between $c_1'(0)$ and $c'_2(0)$ is larger than $\pi/2$. Let $c_3:[0, l]\rightarrow \mathbb{H}^2$ be the geodesic between between $c_1(t_1)$ and $c_2(t_2)$. I have to prove that there exists $C>0$ such that $L(c_3)\geq t_1+t_2 - C$.
In a previous exercise I proved that for such $c_1$ and $c_2$ the angles between $c_1'(t_1)$ and $c_3'(0)$ is no larger than $2e^{-t_1}$ and the angle between $c_2'(t_2)$ and $c_3'(l)$ is no larger than $2e^{-t_2}$.
The hint says to use these fact and the first variation of length formula.
My attempt: Take a point $c_2(s_0)$ and consider the geodesic variation of the geodesic from $c_1(t_1)$ and $c_2(s_0)$, call it $h(r,t)$ for $r\in(-\epsilon, \epsilon)$, $t\in [0,R]$, $R=d(c_1(0), c_2(s_0))$. Then the vector field $X(t)=\partial_r h(r,t)$ is Jacobi field with $X(0)=0$ and $X(R)=c_2'(s_0)$. Therefore by the first variation formula I get $$ \partial_{r=0}L(h(r, \cdot)) = \langle X(R), \partial_t h(0,t)\rangle = \langle c_2'(s_0), \partial_t h(0,t)\rangle.$$ This inner product is the cosinus of the angle between $c_2'(s_0)$ and $\partial_t h(0,t)$, which is smaller than 1. On the other hand, by the estimate explained above this inner product is larger than $\cos(2e^{-s_0})$ and tends to one as $s_0$ tends to infinity. Hence $ \partial_{r=0}L(h(r, \cdot)) \to 1$. I think this implies that $L(h(0,\cdot))$ is in a neigbourhood of the length of $c_1\cup c_2([0,s_0])$, which is $t_1+s_0$, so that for $s_0=t_2$ I have the inequality, but is this true?
Can anyone give me a suggestion? Thanks a lot in advance!