A relationship between subpermanents and the arithmetic-geometric mean inequality (Q958001)

Let \(\mathbf{F}=(f_{ij})\) be a real \(n\times n\) matrix, and let \(1\leq k\leq n\). The \(k\)-subpermanent of \(\mathbf{F}\), denoted by \(\sigma_k(\mathbf{F})\), is defined as follows: Let \({\mathcal D}_{kn}\) denote the set of all subsets of \(\{1,\dots,n\}\) with \(k\) elements. For \(a,b\in{\mathcal D}_{kn}\), let \(\mathbf{F}[a| b]\) denote the \(k\times k\) submatrix of \(\mathbf{F}\) whose row indices are in \(a\) and column indices in \(b\). Then \[ \sigma_k(\mathbf{F})=\sum_{a,b\in{\mathcal D}_{kn}}\text{per } \mathbf{F}[a| b]. \] Assume that \(\mathbf{F}\geq 0\) (entrywise). Denote by \(\mathbf{F}(a| b)\) the submatrix of \(\mathbf{F}\) complementary to \(\mathbf{F}[a| b]\). Define \(\widetilde{\mathbf{F}}=(\widetilde{f}_{ij})\) by \(\widetilde{f}_{ij}=1\) if \(f_{ij}>0\), \(\widetilde{f}_{ij}=0\) if \(f_{ij}=0\). Furthermore, denote \({\mathcal N}=\{(i,j)\mid f_{ij}>0\}\), \({\mathcal S}_k=\{f_{ij}^k| 1\leq i,j\leq n\}, \alpha_{kn}=k!{n\choose k}^2\), and, assuming \(\sigma_k(\widetilde{\mathbf{F}})>0\), \[ \beta_{ij}^{(k)}= \frac{\sigma_{k-1}(\widetilde{\mathbf{F}}(\{i\}| \{j\}))} {k\sigma_k(\widetilde{\mathbf{F}})}. \] Finally, let \(G({\mathcal X})\) and \(A({\mathcal X})\) denote the geometric and respectively arithmetic mean of the elements of a finite set~\(\mathcal X\) of nonnegative real numbers. The authors prove that \[ \alpha_{kn}G({\mathcal S}_k)\leq\sigma_k(\widetilde{\mathbf{F}}) \prod_{(i,j)\in\mathcal N} f_{ij}^{k\beta_{ij}^{(k)}}\leq \sigma_k(\mathbf{F})\leq\sigma_k(\widetilde{\mathbf{F}})\sum_{(i,j)\in\mathcal N}\beta_{ij}^{(k)}f_{ij}^k\leq\alpha_{kn} A({\mathcal S}_k). \] As a corollary, this implies \[ G({\mathcal S}_n)\leq\frac{1}{n!}\text{per }\mathbf{F}\leq A({\mathcal S}_n). \]