Show that $x f \cdot f'' + f \cdot f' - x (f')^2 > 0$ for certain $f(x) = 1 + x - c x^2 + x^3 + x^4$. Consider the polynomial $f(x) = 1 + x - c x^2 + x^3 + x^4$ where $c \ge 0$. Suppose that $|f(z)| < f(|z|)$ for every complex number $z \notin [0, \infty)$. How can we show that $$F(x):= x f(x)\cdot f^{\prime\prime}(x) + f(x) \cdot f^\prime(x) - x (f^\prime(x))^2 > 0$$ for all $0\le x \le 1$? 
 A: The expression that should be positive for $0< x\leq1$ computes to
$$g(x,c):=1+9x^2 +20x^3+9x^4+x^6-cx(4+x+x^3+4x^4)\ .$$
We then can reformulate the question as follows: For which $c\geq0$ do we have
$$p(x):={1+9x^2 +20x^3+9x^4+x^6\over x(4+x+x^3+4x^4)}>c\qquad(0<x\leq1)\ ?$$
This amounts to finding the minimum of $p(x)$ in the given interval. The inherent symmetries of $p$ suggest putting
$$x+{1\over x}=: u\ ,\qquad{\rm resp.}\qquad x={1\over2}\bigl(u-\sqrt{u^2-4}\bigr)\qquad(u\geq2)\ .$$
After some calculations one finds that 
$$q(u):=p\bigl(x(u)\bigr)={u^3+6u+20\over 4u^2+u-8}\qquad(u\geq2)\ .$$
A plot of $q$ shows that $q$ is unimodal in the given range with a global minimum between $u=4$ and $u=5$. The minimum is found by solving $q'(u)=0$, which leads to the third degree equation $u^3-12u-34=0$. Fortunately the unique real solution $u_*$ can be expressed as
$$u_*=\tau+2\tau^2\doteq4.4372\ ,$$
with $\tau:=2^{1/3}$. The minimum of $q(u)$ for $u\geq2$, hence the minimum of $p(x)$ for $0<x\leq1$ is then given by
$$q(u_*)={6(2\tau^2+\tau+3)\over 2\tau^2+11\tau+8}\doteq1.78191\ .$$
It follows that the desired inequality holds when $0\leq c<q(u_*)$.
