Prove $\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1 $ Prove $$\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1,
$$
where $a,b,c \in (0,1)$.
I tried to solve it in this way:
$$\log_{a}{(\frac{b^2}{ac}-b+ac})\geq1$$
$$\frac{b^2}{ac}-b+ac\geq a$$
$$\frac{(b-ac)^2+abc}{ac}\geq a$$
$$\frac{(b-ac)^2}{ac}+b\geq a$$
We ca affirm that
$$\frac{(b-ac)^2}{ac}\geq 1,b \geq1, a\geq 1$$ then the product
$\frac{(b-ac)^2}{ac}+b\geq a$ is true.
Analogous for $\log_{b}{(\frac{c^2}{ab}-c+ab})$ and $\log_{c}{(\frac{a^2}{bc}-a+bc})$.
The question is whether this solution is properly addressed?
 A: Another attempt:
At least of logarithms must be $\geq1$. We suppose the first:
$$\frac{b^2}{ac}-b+ac=ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)$$
$$ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)\geq a\leftrightarrow c\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)-1\geq0$$
Let
$\frac{b}{ac}=x$:
the function $f(x)=c(x^2-x+1)-1=cx^2-cx+c-1$ have the minimum $f_{min}=\frac{3}{4}c-1$, for $x=\frac{b}{ac}=\frac{1}{2}$.
$$f_{min}\geq0\rightarrow \frac{3}{4}c\geq1\rightarrow c\geq\frac{4}{3}>1$$
hence $c\notin(0,1)$!
So, I think, the inequality is not true.
A: Your idea is good but I think it would be more rigorous in the following way:
Before proving
$$F(a,b,c) = \log_{a}\left(\frac{b^2}{ac}-b+ac\right)\cdot\log_{b}\left(\frac{c^2}{ab}-c+ab\right)\cdot\log_{c}\left(\frac{a^2}{bc}-a+bc\right)\geq1$$
one can prove that


*

*$\frac{b^2}{ac}-b+ac \ge b$

*$\frac{c^2}{ab}-c+ab \ge c$

*$\frac{a^2}{bc}-a+bc \ge a$

For example, the first one can be proven as follows (others are the same):
$$\begin{align}
&\frac{b^2}{ac}-b+ac \ge b \\
&\iff\frac{b^2 - 2bac + (ac)^2}{ac} \ge 0 \\
&\iff\frac{(b-ac)^2}{ac}\ge 0 \qquad\square\end{align}$$
Hence,
$$F(a,b,c) \ge \log_ab\cdot\log_bc\cdot\log_ca = \frac{\ln b}{\ln a}\cdot\frac{\ln c}{\ln b}\cdot\frac{\ln a}{\ln c} = 1$$
