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the question is in the title.

I know that this is a direct consequence of Newton's inequalities but I'm looking for a proof without using it. A hint was given to solve it :

Show that

$$ P'^2 - P''P\ge 0$$

To show the given identity, let $$P(X) = \lambda\prod_{k=0}^{r} (X-\alpha_k)^{m_k} = \sum_{k=0}^{n} a_k x^k$$ The identity : $$ \dfrac{P'}{P} = \sum \dfrac{m_k}{X-\alpha_k}$$ Shows that $\dfrac{P'}{P}$ is decreasing, but : $$ \left(\dfrac{P'}{P}\right)' = \dfrac{P''P - P'^2}{P^2}$$ Therefore, $P''P - P'^2\le 0$

To prove that $a_{k-1}a_{k+1}\le a_k^2$ for $0< k<n$, I don't really know how to start. I thought of using Vieta's formulas to express $a_{k-1}a_{k+1}$ and $a_k^2$ but I don't think it will really help me use the fact that $P'^2 - P''P\ge 0$. Any thoughts ?

Thanks for reading.

aleph0
  • 923

1 Answers1

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Here's a solution :

Let $$P(X) = \sum_{k=0}^{n} a_kX^k$$ $$Q(X) = (P'^2 - P''P)(X) = \sum_{k=0}^{2n-2} c_kX^k$$

The first observation is that $c_0 = a_1^2 - 2a_0a_2$. Therefore, if $P$ has real roots, $Q(0)\ge 0$ so that $a_1^2\ge 2a_0a_2$. if $a_0a_2<0$, $a_1^2\ge a_0a_2$, otherwise, $a_0a_2\le 2a_0a_2\le a_1^2$ and the result follows for $k=1$.

For other values of $k$, apply the same reasoning to $P^{(k-1)}$ which also has real roots :

Let $P^{(k-1)} = \sum_{j=0}^{n-k+1} c_j X^j$. We know that $c_1^2\ge 2c_0c_2$.

But $c_0 =(k-1)!\cdot a_{k-1}$, $c_1 = k!\cdot a_k$ and $c_2 =(k+1)!a_{k+1}/2$. So that : $$ k!^2a_k^2\ge (k-1)!(k+1)! a_{k-1}a_{k+1}$$ $$ \implies ka_k^2\ge (k+1)a_{k-1}a_{k+1}$$ $$ \implies a_k^2\ge \dfrac{k+1}{k} a_{k-1}a_{k+1}$$ And the result follows.

aleph0
  • 923