I have a question about the proof of the this statement (page 42 in "Non-positive curvature: geometric and analytic aspect", J. Jost):
Lemma: Let $M$ be a complete, simply connected Riemannian manifold with sectional curvature $K\leq - k <0$. Let $p\in M$ and let $c_1, c_2:[0,1]\rightarrow M$ be geodesics starting at $p$ and let $c:[0,1]\rightarrow M$ be the geodesic from $c_1(1)$ to $c_2(1)$. Then there exists a constant $C(k)>0$ depending on the curvature such that for every $t\in[0,1]$ there exist $s\in[0,1]$ and $i\in\{ 1,2\}$ with $d(c(t), c_i(s)) \leq C(k)$.
They claim this is a consequence of the following result, which I managed to prove with Rauch Comparison Theorem:
Claim: Consider $M$ as in the Lemma take $p\in M$ and $c:[0,1]\rightarrow M$ a geodesic which does not contain $p$. Let $f(s)=d^2(p, c(s))$, then $f''(s)\geq 2\sqrt{k}d(p, c(s))\coth(\sqrt{k}d(p, c(s))\Vert c' \Vert^2$
How does the Lemma follow from the claim?
My attempt is the following: take $c_1, c_2, c$ as in the Lemma. Consider $q$ the footpoint of the perpendicular from $p$ to $c$ so that $d(p, c([0,1]))=d(p,q)$. Fix $t\in [0,1]$, then I think it is possible to find a point $s\in[0,1]$ such that $d(c(t), c_1(s))=d(q,p)\leq d(c(t), p)=: g(t)$. Now we consider a function $f:[0,1]\rightarrow \mathbb{R}$ such that $f(0)=L(c_1)^2, f(1)=L(c_2)^2$ and $f''(t) = 2\sqrt{k}g(t)\coth(\sqrt{k}g(t))\Vert c' \Vert^2$. This should imply that $d(p, c(t)) \leq f(t)$. But then, how can I further estimate $f(t)$ with respect to the curvature?
Definition. Let $X$ be a complete simply connected Riemannian manifold of sectional curvature $\le -k<0$, let $X_{-k}$ be the hyperbolic plane of the (constant) curvature $-k$. Given a geodesic triangle $\Delta xyz$ in $X$, its comparison triangle in $X_{-k}$ is a geodesic triangle $\Delta \bar{x} \bar{y} \bar{z}$ whose respective side-lengths equal those of $\Delta xyz$. Given a point $p$ on the geodesic $yz$ in $X$, its comparison point $\bar{p}\in \bar{y} \bar{z}$ is the unique point such that $$ d_{X}(y,p)=d_{X_{-k}}(\bar{y}, \bar{p}). $$
The claim that you proved implies that every geodesic triangle $\Delta xyz$ in $X$ and its comparison triangle $\Delta \bar{x} \bar{y} \bar{z}$ in $X_{-k}$ satisfy the comparison inequality: $$ d_X(x,p)\le d_{X_{-k}}(\bar{x}, \bar{p}). $$ The reason is that the inequality on the 2nd derivatives implies that the distance function $d_X(x,p(t))$ decreases faster than $d_{X_{-k}}(\bar{x}, \bar{p}(t))$ (until it reaches the point of global minimum), where $t=d_X(x, p(t))$ is the arc-length parameter for the geodesic $xy$.
Subdividing the triangle $\Delta xyz$ along the segment $xp$ and applying the comparison inequality again yields the 2nd comparison inequality: $$ d_X(p,q)\le d_{X_{-k}}(\bar{p}, \bar{q}), $$ for all $q\in xy, p\in yz$ and their comparison points $\bar{q}\in \bar{x}\bar{y}$, $\bar{p}\in \bar{y} \bar{z}$.
Now, what's left is to prove
Lemma 1. Suppose that $\Delta \bar{x} \bar{y} \bar{z}$ is a geodesic triangle in $X_{-k}$. Then every point $\bar{p}\in \bar{y} \bar{z}$ on the side $\bar{y} \bar{z}$ of the triangle belongs to the $C(k)$-neighborhood of the union of the two other sides of the triangle.
One can either use hyperbolic trigonometry or direct calculation in a hyperbolic plane model or the Gauss-Bonnet formula to get such an inequality. For instance $C(k)=\sqrt{2/k}$ will work.
Combined with the second comparison inequality, it yields
Lemma 2. Suppose that $\Delta x y z$ is a geodesic triangle in $X$. Then every point $p\in y z$ belongs to the $C(k)$-neighborhood of the union of the two other sides of the triangle.