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!



Unusual inequality concerning elementary symmetric functions
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.


