Inequality $3 \ln(q) (1-x)(1-qx) + \ln(qx) \geq 0$ for $q^{-2} < x < q^{-1}$ Let $q \geq 2$. I would like to know if 
$$f_q(x) := 3 \ln(q) (1-x)(1-qx) + \ln(qx)$$
is non-negative whenever $x \in ]q^{-2}, q^{-1}[$.
I tried to differentiate $f$ with respect to $x$ or to $q$, but I am stuck. Typically, finding the zeros of 
$f_q'(x) = 3 \ln(q) (2qx - q - 1) + 1/x$ leads to a quadratic equation in $x$ with ugly solutions...
Any help would be appreciated.
 A: We have 
$$f_q'(x)=\frac{g(x)}{x}$$where $g(x)=6q\ln(q)x^2-3(q+1)\ln(q)x+1$ is an upward parabola.
Now, $f_q(x)\ge 0$ for $x\in(q^{-2},q^{-1})$ follows from
$$f_q(q^{-1})=0\tag1$$
$$f_q(q^{-2}) \gt 0\tag2$$
$$f_q'(q^{-1})\lt 0\tag3$$
Proof for $(1)$ : 
$$f_q(q^{-1})=3 \ln(q)\left(1-\frac 1q\right)(1-1) + \ln(1)=0$$
Proof for $(2)$ :
$$f_q(q^{-2})=\frac{h(q)}{q^3}\ln(q)$$ where $h(q)=2q^3-3q^2-3q+3$. Since $h'(q)=6\left(q-\frac{1-\sqrt 3}{2}\right)\left(q-\frac{1+\sqrt 3}{2}\right)$, we see that $h'(q)\gt 0$ and $h(q)\gt 0$ follow from $h'(2)\gt 0$ and $h(2)\gt 0$.
Proof for $(3)$ : 
$$\begin{align}f_q'(q^{-1})&=-(q-1)(3\ln(q)-1)+1
\\\\&\le -(2-1)(3\ln(2)-1)+1
\\\\&=2-3\ln(2)
\\\\&\lt 0\end{align}$$

Added : 
Let $\alpha,\beta\ (\alpha\lt \beta)$ be the roots of $f_q'(x)=0$. Then, we have $\alpha\lt q^{-1}\lt \beta$ from $(3)$. Now, we have two cases :
Case 1 : If $f_q'(q^{-2})\gt 0$, then $f_q(x)$ is increasing for $q^{-2}\lt x\lt \alpha$ and is decreasing for $\alpha\lt x\lt q^{-1}$, so $f_q(x)\ge 0$ follows.
Case 2 : If $f_q'(q^{-2})\le 0$, then $f_q(x)$ is decreasing for $q^{-2}\lt x\lt q^{-1}$, so $f_q(x)\ge 0$ follows.
So, in either case, we have $f_q(x)\ge 0$.
