$\sum_{i=1}^{n-1} \left|\dfrac{a_ia_{n-i}}{a_n}\right| \geq C_{2n}^n-1$ Given that the equation $$p(x)=a_0x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n=0$$ has $n$ distinct positive roots, prove that 
$$\sum_{i=1}^{n-1} \left|\dfrac{a_ia_{n-i}}{a_n}\right| \geq C_{2n}^n-1$$
I had tried to calculate $P'(x)$ but can't go further. Please help me. Thanks
 A: 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
$$
\left|\frac{a_1a_{n-1}}{a_0a_n}\right|
=\left(\sum_{k_1}r_{k_1}\right)\left(\sum_{k_1}\frac1{r_{k_1}}\right)
\ge\binom{n}{1}^2
$$
$$
\left|\frac{a_2a_{n-2}}{a_0a_n}\right|
=\left(\sum_{k_1<k_2}r_{k_1}r_{k_2}\right)\left(\sum_{k_1<k_2}\frac1{r_{k_1}r_{k_2}}\right)
\ge\binom{n}{2}^2
$$
$$
\left|\frac{a_3a_{n-3}}{a_0a_n}\right|
=\left(\sum_{k_1<k_2<k_3}r_{k_1}r_{k_2}r_{k_3}\right)\left(\sum_{k_1<k_2<k_3}\frac1{r_{k_1}r_{k_2}r_{k_3}}\right)
\ge\binom{n}{3}^2
$$
$$
\vdots\tag{2}
$$
Summing $(2)$ yields
$$
\sum_{i=1}^{n-1}\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}-2\tag{3}
$$
If we include the end terms, we get the arguably more aesthetic
$$
\sum_{i=0}^n\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}\tag{4}
$$

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.
A: Hint:


*

*Represent $\dfrac{a_k}{a_0}$ in terms of roots, and try to figure out the relationship between $\dfrac{a_k}{a_0}$ and $\dfrac{a_{n-k}}{a_0}$.

*Use Cauchy-Schwarz Inequality.

*Use the equality $\sum_{p\ge 0} C_n^p C_n^{n-p} = C_{2n}^n$.
