Minimization of $log_{a}(bc)+log_{b}(ac)+log_{c}(ab)$? I am trying to find the minimal value of the expression: 
$log_{a}(bc)+log_{b}(ac)+log_{c}(ab)$
I think experience gives that the variables should be equal, if so then the minimal value is 6, but this not true in general.
Any hints or help will be greatly appreciated. 

Assuming $a > 1$ and $a <= b <= c$.  Let $q = b/a$ and $r = c/b$, so b = $qa$ and $c = qra$, and $ q,r \ge 1$.
$$ f(a,b,c) = \lg_a a^2 q^2 r + \lg_{aq} a^2 q r + \lg_{aqr} a^2 q $$
$$ = \dfrac{2 \ln a + 2 \ln q + \ln r}{\ln a} + \dfrac{2 \ln a + \ln q + \ln r}{\ln a + \ln q} + \dfrac{2 \ln a + \ln q}{\ln a + \ln q + \ln r}$$
$$ = 6 + \dfrac {2 \ln q + \ln r}{\ln a} + \dfrac{-\ln q + \ln r}{\ln a + \ln q} + \dfrac{-\ln q - 2 \ln r}{\ln a + \ln q + \ln r}$$
We know $\ln q$ and $\ln r$ are both $\ge 0$, so $\dfrac {\ln q}{\ln a}$ is >= to both $\dfrac {\ln q}{\ln a + \ln q}$ and $\dfrac {\ln q}{\ln a + \ln q + \ln r}$, let $\epsilon_1 \ge 0$ be the sum of these differences, so
$$ f(a,q,r) = 6 + \epsilon_1 + \dfrac{\ln r}{\ln a} + \dfrac{\ln r}{\ln a + \ln q} + \dfrac {-2 \ln r}{\ln a + \ln q + \ln r}$$
Similarly, both $\dfrac{\ln r}{\ln a}$ and $\dfrac{\ln r}{\ln a + \ln q}$ are greater than $\dfrac {\ln r}{\ln a + \ln q + \ln r}$, and let $\epsilon_2 \ge 0$ be the sum of the differences.
$ f(a,q,r) = 6 + \epsilon_1 + \epsilon_2$, and both epsilons are greater than 0, so $f(a,q,r) >= 6$.  When $q = r = 1$, $f(a,q,r) = 6$.  So 6 is the minimum.
