Can we find a complete noncompact Riemannian manifold $(M^n,g)$ with bounded geometry satisfying the following conditions?
the curvature operator $Rm>0$;
the scalar curvature $R \ge 1$.
Notice that any such manifold must be diffeomorphic to $\mathbb R^n$.
Yes, this is possible. Note that a strictly convex hypersuface in $\mathbb R^{n+1}$ has positive $Rm$.
To get an example, consider the following graph $H\subset \mathbb R^{n+1}$ over the open unit $n$-disk in $\mathbb R^n$:
$$H:=\left(x_{n+1}=\frac{1}{1-\sum_{i=1}^n{x_i^2}}\right).$$
Clearly, $H$ is convex. Moreover, the scalar curvature of $H$ tends to the scalar curvature of the unit $n-1$-sphere as $x_{n+1}$ tends to infinity. In particular the scalar curvature is greater than a certain positive $c>0$ on $H$. So if one scales down this hypersurface by a constant (i.e takes the hypersurface $\varepsilon\cdot H\subset \mathbb R^{n+1}$), we get $R(\varepsilon\cdot H)>1$. The example works if $n\ge 3$.
