Given the vectors $A$ and $B$, and a curve with a tangent vector $U$, I need to show that if $\frac{d}{d\lambda}\left(g_{\mu\nu}A^\mu B^\nu \right)=0$, then $ U^\alpha\nabla_\alpha A^\mu=0 $ and $ U^\alpha\nabla_\alpha B^\mu=0 $.
I've done this going in the other direction, but going in this direction is giving me trouble. I've gotten to the equation $$ -g_{\mu\nu}B^\nu U^\alpha \nabla_\alpha A^\mu=g_{\mu\nu}A^\mu U^\alpha \nabla_\alpha B^\nu $$ But I can't see where to go from here. When I emailed my professor, all he said that because $g_{\mu\nu}$ is symmetric, then that means both sides of my equation are the same, but one has a minus sign, which leads to both sides being $0$. I'm not seeing how the metric being symmetric leads to both sides of the equation being identical, as on one side the derivative is on $A$ and on the other it is on $B$. Any help with this would be greatly appreciated.
Trivial counter-example: on $\Bbb R$, let $A(t) = 0$ and $B(t) = t\frac{\partial}{\partial t}$. Then $\langle A,B\rangle =0$ even though $B$ is not parallel.
Non trivial counter-example: consider $\Bbb R^2$ with its euclidean structure, and $\gamma(t) = (t,0)$, which is a geodesic. Let $A(t) = (\cos t, \sin t)$ and $B(t) = (-\sin t, \cos t)$ be tangent vectors along $\gamma$. Then $\langle A, B\rangle = 0$ even though neither $A$ nor $B$ is parallel.