= Solution
Let
$$
L(x,\dot x)=\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}
=\frac{ds}{d\lambda}.
$$
The <Euler-Lagrange equation> gives
$$
\frac d{d\lambda}\left(\frac{g_{\beta\nu}\dot x^\nu}{L}\right)
-\frac1{2L}\partial_\beta g_{\mu\nu}
\dot x^\mu\dot x^\nu=0.
$$
Since $\dot x^\mu=L\,dx^\mu/ds$ and $d/d\lambda=L\,d/ds$, this becomes
$$
\frac d{ds}\left(g_{\beta\nu}\frac{dx^\nu}{ds}\right)
-\frac12\partial_\beta g_{\mu\nu}
\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0.
$$
Expanding the derivative, multiplying by the inverse metric, and using the <Christoffel symbol>
$$
\Gamma^\alpha_{\mu\nu}
=\frac12g^{\alpha\beta}
(\partial_\mu g_{\beta\nu}
+\partial_\nu g_{\beta\mu}
-\partial_\beta g_{\mu\nu})
$$
gives the <geodesic equation>
$$
\boxed{
\frac{d^2x^\alpha}{ds^2}
+\Gamma^\alpha_{\mu\nu}
\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0}.
$$
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