Prove $\log_ab+\log_bc+\log_ca\geq1+\log_{ab}bc+\log_{bc}ab$ Prove inequality $$\log_ab+\log_bc+\log_ca\geq1+\log_{ab}bc+\log_{bc}ab$$
for $a>1,b>1,c>1.$
Inequality is interesting because of asymmetry and inhomogeneity and I think the solution might interest someone.
We noted $x=\lg a,y=\lg b, z=\lg c $ and wrote inequality in the form
$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\geq 1+\frac{y+z}{x+y}+\frac{x+y}{y+z}$$
for $x>0, y>0, z>0.$
We denote $$ \frac{y}{x}=A,\frac{z}{y}=B, \frac{x}{z}=C$$ with $ABC=1$ and we get
$$A+B+C\geq \frac{A+AB}{1+A}+\frac{1+A}{A+AB}+1.$$
For $C=\frac{1}{AB}$ is obtained
$$\begin{align} A^3B^2+A^3B+AB^3-A^2B^2-2A^2B-2AB+A+1 &\geq0 \\
\implies\quad\quad\quad (A+1)(AB-1)^2+AB(A-B)^2 &\geq0.
\end{align}$$
This is a solution. Has anyone another idea?

We can use C-S.
Indeed, $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}-1-\frac{x+y}{y+z}-\frac{y+z}{x+y}=\frac{x^2}{xz}+\frac{y^2}{yx}+\frac{z^2}{yz}+\frac{y^2}{y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)\geq$
$\geq\frac{(x+y+z+y)^2}{xz+yx+yz+y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)=0$
