Area and Gauss-Bonnet inequalities with scalar curvature

Let $X$ be an $n$-dimensional Riemannian manifold with "large positive" scalar curvature. In this paper, we prove in a variety of cases that if $X$ "spreads" in $(n-2)$ directions {\it "distance-wise"}, then it {\it can't} much "spread" in the remaining 2-directions {\it "area-wise".} Here is a geometrically transparent example of what we plan prove in this regard that illustrates the idea. Let $g$ be a Riemannin metric on $X= S^2\times \mathbb R^{n-2}$, for which the submanifolds \mbox {$\mathbb R_s^{n-2}=s\times \mathbb R^{n-2}\subset X$ and $S^2_y= S^2\times y \subset X$} are {\it mutually orthogonal} at all intersection points $$x=(s,y)\in X=\mathbb R_s^{n-2}\cap S^2_y.$$ (An instance of this is $g=g(s,y)=\phi(s,y)^2ds^2+\psi(s,y)^2dy^2$.) Let the Riemannian metric on $\mathbb R_s^{n-2}$ induced from $(X,g)$, that is $g|_{\mathbb R_s^{n-2}}$, be {\it greater than the Euclidean} metric on $\mathbb R_s^{n-2} =\mathbb R^{n-2}$ for all $s\in S^2$. (This is interpreted as "large spread" of $g$ in the $(n-2)$ Euclidean directions.) {\sf If the {\it scalar curvature of $g$ is strictly greater than that of the unit 2-sphere}, $$Sc(g) \geq Sc(S^2)+\varepsilon=2+\varepsilon, \mbox { }\varepsilon>0,$$ then, provided $n\leq 7$, } (this, most likely, is unnecessary) {\sf there exists a smooth {\it non-contractible} spherical surface $S\subset X$, such that $$area(S)<area(S^2)=4\pi.$$} (This says, in a way, that $(X,g)$ "doesn't spread much area-wise" in the 2 directions complementary to the Euclidean ones.)