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Define $f(x_1, \dots, x_n) = \sum_{l = 2}(l - 1) \sigma_{l}(x_1, \dots, x_n)$, where $\sigma_{l}$ is the $l^{\text{th}}$-elementary symmetric polynomial and $(x_1, \dots, x_n)$ is non-negative. Beyond the obvious bounds \begin{align} \sum_{l = 2}^{n} \sigma_{l}(x_1, \dots, x_n) \leq f(x_1, \dots, x_n) \leq (n-1) \sum_{l = 2}^{n} \sigma_{l}(x_1, \dots, x_n), \end{align} are there sharper inequalities on the sums which can be used to yield something of the sort: \begin{align} A \sum_{l = 2}^{n} \sigma_{l}(x_1, \dots, x_n) \leq f(x_1, \dots, x_n) \leq B \sum_{l = 2}^{n} \sigma_{l}(x_1, \dots, x_n), \end{align} for some constants (or perhaps functions) $A > 1$ and $0 < B < n - 1$? If not, what other inequalities are suitable for this problem?

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No better constants are possible. Taking $x_1=x_2=\cdots=x_n=t$ and sending $t \to 0^{+}$ shows that you can't increase $A$ above $1$, sending $t \to \infty$ shows that you can't lower $B$ below $n-1$. – David Speyer Jun 20 '12 at 16:07

No constants $0<A<B$ or $0<B<n-1$ can be substituted. Consider $x_1=1,x_2=x_3=\cdots=0$. Then the sum is precisely $\sum\limits_{l=2}^n \sigma_l(x_1,\ldots,x_n)$, ruling out any such constant $A$. Now consider $x_1=x_2=\cdots = k$. Then the sum approaches $(n-1)\sum\limits_{l=2}^n \sigma_l(x_1,\ldots,x_n)$ asymptotically, as the smaller symmetric polynomials become negligible as $k$ grows, ruling out any such constant $B$.

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