For any given numbers $\{x_1,...,x_n\}$, the k-th elementary symmetric function is $$ E_k=\sum_{1\le i_1<...<i_k\le n} x_{i_1}x_{i_2}...x_{i_k},~~~~~~~k=1,...,n. $$ I guess the $E_k$ can be control by $E_1$ and $E_n$, namely, there are suitable functions $f,g$ such that $$ g(E_1,E_n) \le E_k \le f(E_1,E_n), ~~~~~~~~~\forall k=1,...,n. $$ But I fail to find suitable functions, so I hope to get help in here. Thanks for any hint or answer.
PS: $x_i$ may not be positive.
The control, about which you say, we can get if any $x_i\geq0$.
If so, let $e_k=\sqrt[k]{\frac{\sum\limits_{1\leq i_1<i_2<...<i_k\leq n}x_{i_1}x_{i_2}...x_{i_k}}{\binom{n}{k}}}.$
Thus, by the Rolle's theorem we can prove that: $$e_1\geq e_2\geq...\geq e_n.$$ From here for any $k\in\{1,2,...,n\}$ we obtain: $$e_1\geq e_k\geq e_n.$$ For $n\geq4$ and real variables $x_i$ there is the following.
Let $e_4^4=\frac{\sum\limits_{1\leq i_1<i_2<i_3<i_4\leq n}x_{i_1}x_{i_2}x_{i_3}x_{i_4}}{\binom{n}{4}}$, $e_3^3=\frac{\sum\limits_{1\leq i_1<i_2<i_3\leq n}x_{i_1}x_{i_2}x_{i_3}}{\binom{n}{3}}$, $e_2^2=\frac{\sum\limits_{1\leq i_1<i_2\leq n}x_{i_1}x_{i_2}}{\binom{n}{2}}$ and $e_1=\frac{x_1+x_2+...+x_n}{n},$ where $e_4^4$ and $e_2^2$ can be negatives.
So the following inequality holds: $$e_4^4\geq4e_1e_3^3-3e_2^4.$$