Need some hints for proving a logarithmic inequality. $$\frac{\log_ax}{\log_{ab}x} +
\frac{\log_b{x}}{\log_{bc}x} +
\frac{\log_cx}{\log_{ac}x} \ge 6$$
Did as you suggested and got this, im stuck again:
$$\log_ab + \log_bc + \log_ca \ge 3$$

$$
\frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b.
$$
Therefore
\begin{align}
& \frac{\log_ax}{\log_{ab}x} + \frac{\log_b{x}}{\log_{bc}x} + \frac{\log_cx}{\log_{ac}x} = (1+ \log_a b) + (1+ \log_b c) + (1+ \log_c a) \\[10pt]
= {} & 3 + (\log_a b + \log_b c + \log_c a) = 3 + \frac{\log b}{\log a} + \frac{\log c}{\log b} + \frac{\log a}{\log c}  = 3 + \frac u v + \frac w u + \frac v w
\end{align}
