The statement in the question is false (as I mention in comments), but $(3)$ seems possibly to be what is meant.
For any positive $\{x_k\}$, Cauchy-Schwarz gives
$$
\left(\sum_{k=1}^nx_k\right)\left(\sum_{k=1}^n\frac1{x_k}\right)\ge n^2\tag{1}
$$
Let $\{r_k\}$ be the roots of $p$, then
$$
\left|\frac{a_1a_{n-1}}{a_0a_n}\right|
=\left(\sum_{k_1}r_{k_1}\right)\left(\sum_{k_1}\frac1{r_{k_1}}\right)
\ge\binom{n}{1}^2
$$
$$
\left|\frac{a_2a_{n-2}}{a_0a_n}\right|
=\left(\sum_{k_1
Note that $(3)$ and $(4)$ are sharp. If we cluster roots near $1$, we will get coefficients near $(x-1)^n$, for which the sums in $(3)$ and $(4)$ are equal to their bounds.