Assume that $(M^n,g)$ is an $n$ dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?
There is no reason for this, and the answer is indeed no. The simplest example I can think of is the product of two $\mathbb{S}^2$, each endowed with the round metric. This manifold is homogeneous and thus has constant scalar curvature, its sectional curvature is nonnegative so its Ricci tensor also is (and is in fact even positive), but the Ricci curvature in a direction $u$ depends on the angle between $u$ and the tangent spaces to the fibers of the projection on each factor (i.e., on whether $u$ is close to be horizontal or vertical or not). 


As an example where this does hold, for $\omega$ a Kähler metric of constant scalar curvature with $\pi c_1(M) = \lambda [\omega]$, then $\omega$ is KählerEinstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry". 

