Let $(M^4,g)$ be a complete four-dimensional Riemannian manifold. First, if the $Q$-curvature $Q_g\geq 6k^2$ and scalar curvature $R_g\geq -12k$ for some positive constant $k$, then $(M^4,g)$ is either Einstein with $Ric_g=-3kg$ or compact with $R_g\ge 12k$. As a corollary, the fundamental group $\pi_1(M^4)$ satisfies $|\pi_1(M^4)|\leq 16\pi^2/(\int_{M^4}Q_g d\mu_g),$ under the additional assumption $R_g>-12k$. Second, if the scalar curvature $R_g>0$ and $Q_g\geq \theta R_g$ for a positive constant $\theta$, then $M^4$ is compact and the diameter of $(M^4,g)$ is at most $4\pi/\sqrt{15\theta}$.
Let $(M^n,g)$ be a connected, closed, smooth Riemannian manifold with dimension $n\geq 3$. There exists a positive constant $\varepsilon_n<1$ such that, if Ricci curvature $\operatorname{Ric}_g\geq \varepsilon_n(n-1)g$ and the scalar curvature $R_g\geq n(n-1)$, then the volume $V_g(M^n)$ is less than or equal to the volume of standard $n$-sphere. This confirms a conjecture by Bray in 1997.