Show that $\log_a(bc)+\log_b(ac)+\log_c(ab)\geq 6$ Show that  if $a,b,c\in(1,\infty)$, then $\log_a(bc)+\log_b(ac)+\log_c(ab)\geq 6$. 
 A: $$\log_a(bc)+\log_b(ac)+\log_c(ab)=\log_a b+ \log_ba+\log_ac+\log_ca+\log_bc+\log_cb$$
By AM-GM
$$log_ab+\log_ba\ge2\sqrt{\log_ab \log_ba}=2$$
And similarly for the other two pairs. Hence we have
$$\log_a(bc)+\log_b(ac)+\log_c(ab)\ge 2 + 2 + 2$$
$$=6$$
A: By Cauchy-Schwarz, for $a,b,c\gt0$, we have
$$
\begin{align}
\frac{b+c}a+\frac{c+a}b+\frac{a+b}c
&=(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)-3\\
&\ge\left(a\cdot\frac1a+b\cdot\frac1b+c\cdot\frac1c\right)^2-3\\[6pt]
&=6
\end{align}
$$
Substitute $a\mapsto\log(a)$, $b\mapsto\log(b)$, and $c\mapsto\log(c)$ to get the desired inequality for $a,b,c\gt1$.
A: Hint. Note that
$$\log_a(b)+\log_b(a)=\frac{\ln(b)}{\ln(a)}+\frac{\ln(a)}{\ln(b)}=t+\frac{1}{t} \geq 2$$
where $t:=\frac{\ln(b)}{\ln(a)}>0$ and we used the inequality $x+y\geq 2\sqrt{xy}$.
A: $$A=\ln a,\space B=\ln b,\space C=\ln c$$
$$\frac{\ln b+\ln c}{\ln a}+\frac{\ln a+\ln c}{\ln b}+\frac{\ln a+\ln b}{\ln c}$$$$=\frac{BC(B+C)+AB(A+B)+AC(A+C)}{ABC}$$$$=\frac{(A^2B+B^2C+C^2A)+(A^2C+C^2B+B^2A)}{ABC}$$$$\ge\frac{3\sqrt[3]{A^3B^3C^3}+3\sqrt[3]{A^3B^3C^3}}{ABC}=\frac{6ABC}{ABC}=6$$
A: Let $ f(x)=x+\frac 1x=f(\frac 1x)$.
for $x>y\geq1$,
$$f(x)-f(y)=(x-y)\frac{ xy-1   }{xy   }\geq0$$
$\implies$ f increasing and
$$(\forall x\geq1)\; \;\; f(x)=f(\frac 1x)\geq f(1)=2$$
$\implies$
$$(\forall x>0)\;\;\; f(x)\geq2$$
thus
$$\forall \; a,b,c\;>0 , \neq 1$$
$$f(\log_a(b))+f(\log_b(c))+f(\log_c(a))\geq6$$
A: Since $$a+1/a > 2 \text{ for all } a > 1$$
and  $$\log_ab = 1 / \log_b a$$
We know that 
$$\log_a b + \log_b a > 2$$
so  $$(\log_a b + \log_b a) + (\log_a c + \log_c a) + (\log_b c + \log_c b) > 6$$
A: 
We show the more general version of this inequality
  $$S=(\log_a^{bc})^r+(\log_b^{ac})^r+(\log_c^{ab})^r\geq 3\times 2^r$$

we can write $\log_a^{bc}=\frac{\log bc}{\log a}=\frac{\log b}{\log a}+\frac{\log c}{\log a}\geq 2\sqrt{\frac{\log b\log c}{\log^2a}}$, so 
$$(\log_a^{bc})^r\geq \frac{2^r(\log b\log c)^{\frac{r}{2}}}{(\log a)^r}$$
Hence $$S\geq \frac{2^r(\log b\log c)^{\frac{r}{2}}}{(\log a)^r}+\frac{2^r(\log a\log c)^{\frac{r}{2}}}{(\log b)^r}+\frac{2^r(\log a\log b)^{\frac{r}{2}}}{(\log c)^r}$$
again by using geometric means inequality we have $$\frac{S}{3}\geq \left[\frac{2^r(\log b\log c)^{\frac{r}{2}}}{(\log a)^r}\times\frac{2^r(\log a\log c)^{\frac{r}{2}}}{(\log b)^r}\times\frac{2^r(\log a\log b)^{\frac{r}{2}}}{(\log c)^r}\right]^{\frac{1}{3}} $$
Hence $$S\geq 3\left[2^{3r}\frac{(\log a\log b\log c)^r}{(\log a\log b\log c)^r}\right]^{\frac{1}{3}}$$
Hence $S\geq 3\times 2^r$. 

In general, we can show the following inequality by the same proof and
  induction on $n$.
$$S=\sum_{i=1}^{n}(\log_{a_i}a_1a_2...a_{i-1}a_{i+1}...a_n)^r\geq
 n(n-1)^r$$

