Consider the Ricci curvature satisfying $Ric(v,v)>0$ for $v\neq 0$. We know that Ricci curvature can be computed as the sum of sectional curvatures of the planes containing $v$ so that if the sectional curvature $K$ satisfies $K>0$ then necessarily $Ric>0$. But, is the converse true?
The reason I ask is we can have manifolds of positive scalar curvature $R>0$ which do not have $Ric>0$. This can be done by taking a product manifold of say a manifold of positive sectional and negative sectional curvatures (so take product of $S_{r}^{n}$ the n-sphere of radius $r$ and $H^n$, the hyperbolic plane with the standard metrics). Then the Riemann tensor of the product, decomposes as the sum of Riemann tensors of the factors, and taking the trace, so does the Ricci tensor. So taking vectors tangent to $S_r^n$ we get a positive value for $Ric$ and taking vectors tangent to $H^n$ we get negative $Ric$. On the other hand, if we take another trace, we see that the scalar curvature is the sum of the scalar curvatures of the factors, so depending on how we choose $r$ we can have $R>0$, $R<0$, or $R=0$.
However, this trick of using product manifolds doesn't work when comparing Ricci and sectional curvatures.
So, my question is, does $Ric>0$ imply that $K>0$ ($K\geq 0$)? If so how would I prove it? If not, is there some simple counterexample?