To state the question and fix conventions I will introduce some notation from e.g. (Lin-Trudinger, Bull. Aust. Math. Soc. 1994, ``On some inequalities for elementary symmetric functions")

Given $\lambda_1, \dots, \lambda_n$ let

\begin{align*} \sigma_k(\lambda) = \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \lambda_{i_1} \lambda_{i_2} \dots \lambda_{i_k} \end{align*} We will always consider this function as restricted to the ``positive cone" \begin{align*} \Gamma_k = \left\{ \lambda \in \mathbb R^n\ |\ \sigma_j(\lambda) > 0 \mbox{ for all } 1 \leq j \leq k \right\} \end{align*}

Furthermore, let \begin{align*} \sigma_{k;i}(\lambda) = \sigma_k(\lambda)_{| \lambda_{i} = 0}, \end{align*} i.e. the function $\sigma_k(\lambda)$ after replacing $\lambda_i = 0$. To see some familiar facts with this notation, note the Newton inequalities \begin{align*} \sigma_k(\lambda) \sigma_{k-2}(\lambda) \leq \frac{(k-1)(n-k+1)}{k(n-k+2)} [\sigma_{k-1}(\lambda)]^2, \end{align*} as well as the Maclaurin inequality for $\lambda \in \Gamma_k$, $k \geq l \geq 1$. \begin{align*} \left[ \frac{1}{n \choose k} \sigma_k(\lambda)\right]^{\frac{1}{k}} \leq \left[ \frac{1}{n \choose l} \sigma_l(\lambda)\right]^{\frac{1}{l}} \end{align*} One can observe the further elementary identities \begin{align*} \sigma_k(\lambda) =&\ \sigma_{k;i}(\lambda) + \lambda_i \sigma_{k-1;i}(\lambda),\\ \sum_{i=1}^n \sigma_{k;i}(\lambda) =&\ (n-k) \sigma_k(\lambda). \end{align*}

Now to state the actual question. Fix $n = 2k$. Given $\lambda = (\lambda_1, \dots, \lambda_n)$, let \begin{align*} A_{\lambda} =&\ \frac{1}{n-2} \left[ \lambda - \frac{\sigma_1(\lambda)}{2(n-1)} (1,\dots,1) \right] \end{align*} I now assume I have a vector $\lambda$ such that $A_{\lambda} \in \Gamma_k$ (recall $k = \frac{n}{2}$). The inequality I seek is that for all $1 \leq i \leq n$, \begin{align*} (n-1) \sigma_k(A_{\lambda}) \leq \lambda_i \sigma_{k-1;i}(A_{\lambda}). \end{align*} A few remarks. First, if helpful it can be expressed purely in terms of the vector $A_{\lambda}$, by using the formula $\lambda = (n-2) A_{\lambda} + \sigma_1(A_{\lambda}) (1,\dots,1)$. Second, it is known that the condition $A_{\lambda} \in \Gamma_k$ implies that $\lambda_i \geq 0$ for all $i$. Third, I have added the tag, ``differential geometry'' only because this question originally comes from a geometric problem, where $\lambda$ represents eigenvalues of the Ricci tensor and $A_{\lambda}$ is the Schouten tensor. Lastly, it is a fairly straightforward matter to verify that the inequality is true for $n=2,4$. Any help is greatly appreciated!

  • $\begingroup$ In other words, if, say, $i=n$ and $A_{\lambda}=(x_1,\dots,x_n)$ we have to prove $(n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1})$, this is standard if $(x_1,\dots,x_{n-1})\in \Gamma_k$, but the problem is that we are not given this. $\endgroup$ Mar 10, 2016 at 8:26
  • 1
    $\begingroup$ Is there a $k$ that should be an $l$ in Maclaurin's inequality? $\endgroup$ Mar 10, 2016 at 12:22

1 Answer 1


We use the following trick, which is standard in such questions. For numbers $x_1,\dots,x_n$ consider the polynomial $F(t)=\prod_{i=1}^n (t-x_i)$, let $F'(t)=n\prod_{i=1}^{n-1}(t-y_i)$ be its derivative ($y$'s are real provided that $x$'s are --- by Rolle theorem). Note that normalized elementary symmetric functions do not change their values if we replace $x$'s to $y$'s: if we denote $${\tilde \sigma}_m(x_1,\dots,x_n)=\frac1{\binom{n}{m}}\sigma_m(x_1,\dots,x_n),$$ we have $${\tilde \sigma}_m(x_1,\dots,x_n)={\tilde \sigma_m}(y_1,\dots,y_{n-1})$$ for all $m=1,2,\dots,n-1$. This immediately follows from taking derivative and applying Vieta's theorem. In proving inequalities for sigmas it is helpful to differentiate polynomial $f$ several times.

Denote $A_{\lambda}=(x_1,\dots,x_n)$ and suppose that $i=n$. Then expressing $\lambda_n=(n-1)x_n+\sigma_1(x_1,\dots,x_{n-1})$ and substituting 'elementary identity' $\sigma_k(A_{\lambda})=x_n\sigma_{k-1;n}(A_{\lambda})+\sigma_k(x_1,\dots,x_{n-1})$ we get equivalent form of your inequality: $$ (n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1}), $$ which may be rewritten as $$ {\tilde \sigma}_k(x_1,\dots,x_{n-1})\leqslant {\tilde \sigma}_1(x_1,\dots,x_{n-1})\cdot {\tilde \sigma}_{k-1}(x_1,\dots,x_{n-1}). $$

Denote $f(t)=(t-x_1)(t-x_2)\dots (t-x_{n-1})$, $g(t)=f^{(k-1)}(t)$, so $g$ is a polynomial of degree $k$ with $k$ real roots $u_1,\dots,u_k$. We have to prove $$k^2u_1\dots u_k\leqslant (u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k).\,\,\,\,(1)$$ (1) is obvious if all $u_i$ are non-negative, so suppose that $u_k<0$.

We are given that $f(t)(t-x_n)=t^{n}-c_1t^{n-1}+c_2t^{n-2}+\dots+(-1)^k c_k t^{n-k}+\dots$ for positive $c_1,\dots,c_k$, and $f(t)$ has only real roots. Taking $k$-th derivative we see by Rolle theorem that $(f(t)(t-x_n))^{(k)}$ has only real roots and they must be positive. So, $(t-x_n)g'(t)+kg(t)$ has $k$ positive roots. Thus $(g(t)\cdot (t-x_n)^k)'$ has a root $x_n$ of multiplicity $k-1$ and $k$ positive roots.

If $x_n\leqslant 0$, then by Rolle theorem polynomial $(g(t)\cdot (t-x_n)^k)'$ has a root between $x_n$ and $u_k$ (or has roots $x_n$ of multiplicity at least $k$ if $x_n=u_k$.) It contradicts to above observations. Analogously, if $x_n\geqslant 0$ but $u_i\leqslant 0$ for some $i\ne k$, we get a negative root of $(g(t)\cdot (t-x_n)^k)'$, that again contradicts to above.

So, $x_n>0$ and $u_1,\dots,u_{k-1}>0$. Since $W(t):=kg(t)+(t-x_n)g'(t)$ has $k$ positive roots and positive leading coefficient, we have $W(0)(-1)^{k}>0$. On the other hand, $g(0)(-1)^k=u_1\dots u_k<0$. Thus $(-1)^{k-1}x_ng'(0)>0$, i.e. $\sigma_{k-1}(u_1,\dots,u_k)>0$. It easily implies that $u_1+\dots+u_k>0$, thus $(u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k)>0>k^2u_1\dots u_k$ as desired.

  • $\begingroup$ It is nothing but taking derivative and applying Vieta. But very useful! $\endgroup$ Mar 10, 2016 at 18:46
  • $\begingroup$ no, I do not care on how lambdas are ordered, that is why $i=n$ is generic case. $\endgroup$ Mar 11, 2016 at 4:46
  • $\begingroup$ We express everything via $A_\lambda$, as you suggest in the end of your post. $\lambda_n=(n-1)x_n+x_1+\dots+x_{n-1}$, substitute this and use your first 'elementary identity' $\endgroup$ Mar 11, 2016 at 5:04

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