TITLE: A criterion for real-rooted polynomials with nonnegative coefficients
QUESTION [12 upvotes]: Let $P \in \mathbb{R}[X]$, with $\deg P = n$. Is it true that
$P$ has only real roots $\quad \Longleftrightarrow \quad P\cdot P'' + (\frac{1}{n}-1)P'^2 \leq 0$ ?
The direct implication can be shown by using Cauchy-Schwarz inequality, but I still lack the reverse (if true).

I apologize to those who have already answered, but a new condition arise: what if $P$ has only nonnegative coefficients?

REPLY [19 votes]: Also simply for $P(x)=x^4-1$ we have 
$$P\cdot P''+\left(\frac{1}{4} -1 \right)\left(P'\right)^2=-12x^2\leq 0$$
and $P$ has imaginary roots.

The stronger version is also false. Take $P(x)=x^5-x$ which satisfies
$$P\cdot P''+\left(\frac{1}{5} -1\right)\left(P'\right)^2=-12x^4-\frac45<0$$
yet it has complex roots.