= Solution
Let $\tau>0$ be the first conjugate time and choose a nonzero <Jacobi field> $J$ with $J(0)=J(\tau)=0$. It is normal by part 1(b), and $D_tJ(\tau)\ne0$, since zero value and derivative there would force $J=0$. Fix any $T>\tau$ and extend $J$ by zero to a continuous piecewise smooth <vector field> $V$ on $[0,T]$.
For endpoint-vanishing fields, the <Riemannian index form> is
$$
I(X,Y)=\int_0^T\bigl(\langle D_tX,D_tY\rangle-\langle R(X,\dot\gamma)\dot\gamma,Y\rangle\bigr)\,dt.
$$
Integration by parts on $[0,\tau]$ and the <Jacobi field> equation give $I(V,V)=0$ and
$$
I(V,W)=\langle D_tJ(\tau),W(\tau)\rangle.
$$
Choose a smooth normal field $W$, zero at $0,T$, with $W(\tau)=-D_tJ(\tau)$. Then
$$
I(V+\varepsilon W,V+\varepsilon W)=-2\varepsilon|D_tJ(\tau)|^2+\varepsilon^2I(W,W)<0
$$
for sufficiently small positive $\varepsilon$. The corner in $V$ causes no difficulty: smooth endpoint-vanishing fields approximate $V+\varepsilon W$ in the <Sobolev space> $H^1$, and the <Riemannian index form> is continuous in this norm on the compact segment. A smooth approximant therefore still has negative index form.
Realize this smooth field as a fixed-endpoint curve variation using the <Riemannian exponential map>. The allowed <second variation of geodesic energy> gives $E'(0)=0$ and $E''(0)<0$, so some nearby curve has energy less than $T/2$, the energy of the unit-speed reference <geodesic>. The <Cauchy-Schwarz inequality> bounds its length by
$$
L^2\leq T\int_0^T|\dot c|^2\,dt=2TE<T^2.
$$
It is shorter than the reference segment. Thus \b[a geodesic cannot minimize beyond its first conjugate time]. This argument deliberately uses $T>\tau$; it does not rule out minimization up to the <conjugate point> itself.
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