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:

$$\log_ab + \log_bc + \log_ca \ge 3$$

• If you don't add your thoughts respectively specify your question, it is likely to be disliked or even closed. – Redundant Aunt Oct 21 '16 at 16:43
• $$\frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b.$$ Therefore $$\frac{\log_ax}{\log_{ab}x} + \frac{\log_b{x}}{\log_{bc}x} + \frac{\log_cx}{\log_{ac}x} = (1+ \log_a b) + (1+ \log_b c) + (1+ \log_c a)$$ $$= 3 + (\log_a b + \log_b c + \log_c a) = 3 + \frac{\log b}{\log a} + \frac{\log c}{\log b} + \frac{\log a}{\log c} = 3 + \frac u v + \frac w u + \frac v w$$ – Michael Hardy Oct 21 '16 at 17:01

Let $m = \log_x a, n = \log_x b, p = \log_x c\implies LHS = 3 + \dfrac{n}{m}+\dfrac{p}{n}+ \dfrac{m}{p} \ge 3+3 = 6$ by applying AM-GM inequality.
$$\log_{a}{x}=\frac{\log_{b}{x}}{\log_{b}{a}}$$ $$\log_{q}{a}+\log_{q}{b}=\log_{q}{ab}$$ $$\log_{b}{ab}=1+\log_{b}{a}$$
If you write $\frac{\log_a x}{\log_{ab} x}=\frac{\frac{\ln x}{\ln a}}{\frac{\ln x}{\ln ab}} = \frac{\ln ab}{\ln a} = 1+\frac{\ln b}{\ln a}$, and something similar with the other fractions, you can the use the familiar inequality : $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge3$$ which can be derived easily from the inequality between arithmetic and geometric means.
$$\frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b.$$ 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}