Let \((M,g)\) be an \(n\)-dimensional Riemannian manifold (\(n>2\)) and let \(S_g\) be the Schouten tensor of \(g\), given by \[ S_g=\frac{1}{n-2}\left(\mathrm{Ric}-\frac{R}{2(n-1)}\cdot g\right). \] Then, for an integer \(k\) with \(1\leq k\leq n\), the \(k\)-scalar curvature of \(M\) can be defined by \[ \sigma_k(g):=\sigma_k(\Lambda_g), \] where \(\sigma_k\) is the elementary symmetric function in \(\mathbb R^n\) and \(\Lambda_g\) is the set of eigenvalues of the matrix \(g^{-1}\cdot S_g\). In particular, \[ \sigma_1(g)=\frac{R}{2(n-1)} ,\quad \sigma_2(g)=\frac{1}{2(n-2)^2}\left(-|\mathrm{Ric}|^2+\frac{n}{4(n-1)}R^2\right). \] Recently, Y. Ge and G. Wang [Proc. Am. Math. Soc. 140, No. 3, 1041–1044 (2012; Zbl 1238.53026)] proved a nice geometric inequality on 4-dimensional closed Riemannian manifolds with non-negative scalar curvature involving \(\sigma_1(g)\) and \(\sigma_2(g)\) as follows: \[ \frac{8}{3}\mathrm{vol}(g)\int_M\sigma_2(g)dv(g)\leq\left(\int_M\sigma_1(g)dv(g)\right)^2. \] The proof is based on the fact that \(\int_M\sigma_2(g)dv(g)\) is a conformal invariant, a fact which is only true on 4-dimensional manifolds. When the dimension is greater than 4, such an inequality does not hold. In the paper under review, the authors define a new conformal invariant and prove the following 3-dimensional analog of the above inequality.
{Theorem.} Any closed 3-dimensional manifold \(M\) with a metric \(g\) of non-negative scalar curvature satisfies \[ 3\mathrm{vol}(g)\int_M\sigma_2(g)dv(g)\leq\left(\int_M\sigma_1(g)dv(g)\right)^2. \] Equality holds if and only if \((M,g)\) is a space form.