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. 
 A: you can assume $a,b,c>1$ or $a,b,c<1$, it has min.
in case one of $a,b,c <1$, and one of $a,b,c>1$ ,there is no min as it can be $ \to -\infty$
A: 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.
A: You can rewrite this as
$$
\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+\left(z+\frac{1}{z}\right)
$$
where $x=\log_{a}b$, $y=\log_{b}c$, and $z=\log_{c}a$.  (These three variables are not independent: there is the additional constraint that $xyz=1$.)  Each of the three identical terms has a minimum value of $2$, which it assumes at $x=1$ (resp. $y=1$ or $z=1$).  So the whole expression can't be smaller than $6$, which can be achieved by simply letting $a=b=c$.
