Newton's inequalities say that if $f(x) = \sum \binom{n}{k} a_k x^k$ is a polynomial with all real roots then $ a_k^2 > a_{k-1}a_{k+1}$.

The proof this result uses that if $f(x)$ has all real roots, so do $df/dx$ and $x^{\deg f} f(1/x)$. So we found two operations which preserve the class of all-real roots polynomials. Then we can reduce to a quadratic.

Proving $df/dx$ has all real roots use Rolle's theorem, since

$$ f'(\xi) = \frac{f(x_{k+1})-f(x_k)}{x_{k+1}-x_k} = 0$$

can be applied for $k = 1, \dots, n-1$.

I noticed the converse is not true. Newton's inequalities does not imply all real roots (though I can't think of counterexample).

Can we improve Newton's inequalities by using one of the Taylor formulas?

$$ f(x) = f(0) + x \, f'(0) + \tfrac{x^2}{2}f''(\xi) \hspace{0.25in} \text{for}\hspace{0.25in} 0 < \xi < 1$$

Possibly the notion of discriminant $(r_1 - r_2)^2$ should also have to be generalized, and I don't know any real algebraic geometry for that.

In a paper by Jim Pitman it is stated without proof that $f(x)$ has all real roots if $b_k = \binom{n}{k} a_k$ is a Polya frequency sequence (e.g. See Karlin). I wish to understand better the connection to total positivity.

log concavity/strong unimodality). But what is the question? Log concavity by itself says very little about the locations of the roots (e.g., if $f$ is a real polynomial with no positive real zeros and $f(1)> 0$, then there exists $n$ such that $(1+x)^n f$ is strongly unimodal---included is that the product has no negative coefficients, too). $\endgroup$restrictions does real roots place on the coefficients? Can we get a nice elementary proof using Rolle's theorem or Mean Value Theorem? $\endgroup$