using Jacobi identity as mentioned in one of the answers, its very easy to prove the Bianchi identity

\begin{align}
[D_\mu,[D_\nu,D_\lambda]]\psi &= [D_\mu,ig F_{\nu\lambda}] \psi \\
&= D_\mu (F_{\nu\lambda}\psi) - F_{\nu \lambda}D_\mu \psi \\
&= D_\mu F_{\nu\lambda}\psi + F_{\nu \lambda}D_\mu\psi - F_{\nu\lambda}D_\mu\psi \\
&= D_\mu F_{\nu\lambda}\psi \, .
\end{align}
Therefore, 
$$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\lambda F_{\mu\nu} = 0$$
which proves the desired result.