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?
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.
It seems that inverse fails in general.
In terms of roots $z_1,\dots,z_n$ your inequality $P\cdot P'' + (\frac{1}{n}-1)P'^2 \leqslant 0$ may be rewritten as $\sum_{1\leqslant j<k\leqslant n} (\frac1{x-z_j}-\frac1{x-z_k})^2\geqslant 0$ for real $x$. Try a polynomial with roots $i,-i$, $1$ with multiplicity $m$, $0$ with multiplicity $m$. Then, multiplying by $(x-1)^2x^2$ we get $m^2+O(m)$, since any expression $x(x-1)(\frac1{x-z_j}-\frac1{x-z_k})$ is obviously uniformly bounded on the whole real line. But $m^2+O(m)$ guy is positive if $m$ is large enough.
