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.
