Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.