Why do manifolds with negative sectional curvature not have conjugate points?

Let $\gamma:[0,\epsilon]\rightarrow M$ be a geodesic curve, $\epsilon>0$ and $J$ be a Jacobi vector field along $\gamma$ with $J(0)=J(\epsilon)=0$.

I remind you that :

1. $J$ being a Jacobi vector field means that $\dfrac{D^2}{dt^2}J+R(\gamma',J)\gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.
2. $\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right] \kappa(X,Y)=\langle R(X,Y)X,Y\rangle$ for any $X,Y\in TM$ with $\kappa$ the sectional curvature.

Now look at the map $\phi:t\mapsto \|J(t)\|^2$. Its second derivative is easy to compute and using the two remarks you get : $$\begin{align} \phi''(t) & = \langle \dfrac{D^2}{dt^2}J,J\rangle + \langle \dfrac{D}{dt}J,\dfrac{D}{dt}J\rangle\\ & = -\langle R(\gamma',J)\gamma',J\rangle + \|\dfrac{D}{dt}J\|^2 \\ & = -\left[\|X\|^2\|Y\|^2-\langle X,Y\rangle^2\right]\kappa(\gamma',J) + \|\dfrac{D}{dt}J\|^2.\end{align}$$

In particular, if all sectional curvatures are non-positive then $\phi''\geq 0$. So $\phi$ is a convex map and since $\phi(0)=\phi(\epsilon)=0$, $\phi(t)=0$ for any $t\in [0,\epsilon]$. It follows that $J$ has to be trivial.

So we can't have conjugate points in $M$.

This is very late, but to provide an alternative proof:

The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $\mathbb{R}^m$.

Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $p\in M$, and that $\gamma:\left[0,T\right]\to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $\gamma$ with $J(0)=0$ and $||J'(0)||>0$.

Let $\tilde\gamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $\mathbb{R}^m$. Let $\tilde{J}$ be the Jacobian vector field along $\tilde\gamma$ satisfying $\tilde{J}(0)=0$, $\tilde{J}'(0)=(<\gamma'(0),J'(0)>,(||J'(0)||^2-<\gamma'(0),J'(0)>^2)^\frac{1}{2},0,0,...,0)$. Then $J(0)=0=\tilde{J}(0)$, $||\tilde{J}'(0)||=||J'(0)||$, and $<\tilde{\gamma}'(0),\tilde{J}'(0)>=<\gamma(0),J'(0)>$.

For all $t\in\left[0,T\right]$, $max(K(\gamma'(t),X))\leq0=min(\tilde{K}(\tilde{\gamma}'(t),\tilde{X}))$ and $dim(\mathbb{R}^m)\leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||\geq||\tilde{J}(t)||$ for all $t\in\left[0,T\right]$. In particular, since $\mathbb{R}^m$ has no conjugate points, $\tilde{J}'(0)\not=0$ guarantees that $||\tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $\gamma$.

Since $p$ and $\gamma$ were arbitrary, $M$ has no conjugate points.