I was reading the set of notes for my course in Classical Dynamics and I came across the following statement that I have no idea how to show, any help is much appreciated.
Consider a particle described by a set of generalised coordinates $q^a$ where $a = 1,2,3,\cdots,n$. Let $L$ be a purely kinetic Lagrangian, i.e. $L$ has no potential energy term. Then the most general or of $L$ is $L = \frac{1}{2}g_{ab}(q_c)\dot{q}^a\dot{q}^b$ where $g_{ab}$ is an invertible symmetric matrix function of $q$ that depends on the generalised coordinates. E.g. in the Cartesian case we have $g_{ab}(q_c) = \delta_{ab}$.
Then it is a simple exercise to see that the Euler-Lagrange Equation for $L$ is $\ddot{q}^a + \Gamma^a_{bc}\dot{q}^b\dot{q}^c = 0$ where $\Gamma^a_{bc} = \frac{1}{2}g^{ad}(\frac{\partial g_{bd}}{\partial q^c}+\frac{\partial g_{cd}}{\partial q^b}-\frac{\partial g_{bc}}{\partial q^d})$.
Unfortunately I don't see how we get the Euler-Lagrange equation from that definition of the Lagrangian. Any hints or solutions much appreciated, especially as I don't think I have quite got the hang of tensors yet.