Given a real polynomial $P(x)$ all whose roots are real, it is not hard to show that $$P(x)P''(x) \leq P'(x)^2 \, \, \, \, (1).$$
Proof sketch: Assume that $P(x) = \prod_{i=1}^n (x-r_i)$. Look at $\log P(x) = \sum_{i=1}^n \log (x-r_i)$ and take the second derivative of both sides.
One nice consequence of Inequality 1, is that if a polynomial $P(x)= a_n x^n + a_{n-1}x^{n-1} \cdots + a_2x^2 + a_1x + a_0$ satisfies $2a_0a_2 > a_1^2 $ then $P(x)$ must have a complex root. To see this plug $x=0$ into the earlier inequality.
One might hope for a version of Inequality 1 which is true for all polynomials with the left-hand side just a function of $P(x)$ and $P''(x)$, and the right just a function of $P'(x)$, that works for all real numbers. But this cannot possibly be true because $P'(x)$ has no information about the constant term of $P(x)$. However, it is true tat Inequality 1 is true for any polynomial provided $|x|$ is sufficiently large.
It also isn't too hard to show that if $Q(x)$ is a quadratic polynomial that
$$Q(x)Q''(x) \leq \left(1+Q'(x)^2 + Q(0)^2\right)^2 \, \, \, \, (2).$$
Together, Inequalities 1 and 2 motivate the following two questions.
- Is there a function $f(n)$ such that for any real polynomial $P(x)$ of degree $n$, one has
$$P(x)P''(x) \leq (1+P'(x)^2 + P(0)^2)^{f(n)}. $$
Edit: This is trivially false for silly reasons as pointed out below. To make it non-trivial, here is question 1a)
1a) Is there a function $f(n)$ such that for any real polynomial $P(x)$ of degree $n$, one has
$$P(x)P''(x) \leq (2+P'(x)^2 + P(0)^2)^{f(n)}. $$
(Here 2 can be replaced with any number greater than 1 with a corresponding increase in $f(n)$).
My guess is that the answer to 1a is still no, but if there is such a function how slow a function can we make $f(n)$?
- Is there a function $f(n)$ such that the following sort of inequality holds for any real polynomial of degree $n$:
$$P(x)P''(x) \leq \left(1+ P'(x)^2 + \sum_{i=0}^{f(n)} P(i)^2\right)^2.$$.
My guess here is that this should be the case, and that one can actually even have $f(n)$ here be linear, maybe something like $n+1$.
