Let $(M,g)$ be a Riemannian manifold. I am trying to show that
\begin{align*} \text{div}(G)\leq 0 \end{align*}
where $G$ is the Einstein $(1,1)$-tensor given by
$$ G=\operatorname{Rc}-\frac{\operatorname{sc}}{2} \operatorname{Id} $$ where $\operatorname{Rc}$ is the Ricci $ (1,1)$-tensor given by $g(\operatorname{Rc}(X),Y)=\operatorname{Ric}(X,Y)$ and $\operatorname{sc}$ is the scalar-curvature. The divergence of a $(k,1)$-tensor $T$ is given as
\begin{align*} \text{div}(T)(X_1,\dots,X_k)=\text{Tr}(Y\mapsto \nabla_YT(X_1,\dots,X_k)) \end{align*}
where $\nabla$ denotes the Levi-Civita connection.
I am a little lost at what to do. The only thing I can think of is to pick a orthonormal basis of $\mathfrak{X}(M)$ and then use the metric to compute the trace. In doing this I tried using Koszul's formula to expand $\nabla_{E_i}G$, however I just end up with some terms involving Lie-brackets of the form $[\operatorname{Rc}(X),E_i]$ which I can't seem to get rid of.
