Find the minimum value of $S$

Question

Let $a,b,c$ be real numbers greater than 1.

Let $S=\log_a{bc}+\log_b{ac}+\log_c{ba}$

Then find the minimum value of $S$

Answer 1

First, we divide the logs of the multiplications into sums:

$\log_a{bc}+\log_b{ac}+\log_c{ba}=\log_a{b}+\log_a{c}+\log_b{a}+log_b{c}+\log_c{a}+\log_c{b}$

After that, we use the $log_a{b}=\frac{\ln{a}}{\ln{b}}$ to further deconstruction:

$\log_a{b}+\log_a{c}+\log_b{a}+log_b{c}+\log_c{a}+\log_c{b}=\frac{\ln{b}}{\ln{a}}+\frac{\ln{c}}{\ln{a}}+\frac{\ln{a}}{\ln{b}}+\frac{\ln{c}}{\ln{b}}+\frac{\ln{a}}{\ln{c}}+\frac{\ln{b}}{\ln{c}}$

Because a, b and c are reals >1, their natural logarithm is a real > 0.

Substituting $x:=\frac{\ln{a}}{\ln{b}}$, $y:=\frac{\ln{a}}{\ln{c}}$, $z:=\frac{\ln{b}}{\ln{c}}$, we are looking for the minimum of $x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}$.

The minimum of a number plus its recipe is 2 for positive real numbers [prove is trivial, I give you on need], thus the minimum is $\underline{\underline{6}}$.

P.s. also trivial that we get this minimum if $a=b=c$.