$\sum_{i=1}^{n-1} \left|\dfrac{a_ia_{n-i}}{a_n}\right| \geq C_{2n}^n-1$

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

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What is $C^n_{2n}$? The central binomial coefficient? – Mike Spivey Dec 17 '12 at 20:05
I guess $C_n^p={n\choose p}$. – Mercy King Dec 17 '12 at 20:07
use Cauchy-Schwarz inequality. – Yimin Feb 26 '13 at 23:17
This doesn't make sense. If we replace all the coefficients by $\lambda a_k$, where $\lambda\gt0$, then $$p(x)=\lambda a_0x^n+\lambda a_1x^{n-1}+\dots+\lambda a_{n-1}x+\lambda a_n=0$$ and yet $$\sum_{i=1}^{n-1}\left|\frac{\lambda a_i \lambda a_{n-i}}{\lambda a_n}\right|=\lambda\sum_{i=1}^{n-1}\left|\frac{a_i a_{n-i}}{a_n}\right|$$ can be made as small as we wish. Did you mean $a_n^2$ in the denominator? – robjohn Mar 9 '13 at 6:48
Even $a_n^2$ in the denominator doesn't work. For any given $p$ and $\lambda\gt0$, we have another $$p_\lambda(x)=a_0x^n+\lambda a_1x^{n-1}+\dots+\lambda^{n-1}a_{n-1}x+\lambda^na_n=0$$ whose roots are $\lambda$ times the roots of $p$ (hence positive and distinct), yet $$\sum_{i=1}^{n-1}\left|\frac{\lambda^ia_i\lambda^{n-i}a_{n-i}} {\lambda^{2n}a_n^2}\right|=\frac1{\lambda^n}\sum_{i=1}^{n-1}\left|\frac{a_ia_{n-‌​i}}{a_n^2}\right|$$ can be made any size we wish. I think these types of scaling can be discounted if the denominator is $a_0a_n$. – robjohn Mar 9 '13 at 7:23

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.

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Hint:

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

2. Use Cauchy-Schwarz Inequality.

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

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