Let $ ln(a) = x$ ,$ ln(b) = y$ and $ ln(c) = z$ ,$ x,y,z > 0$ so the inequality to prove is \[ \sum \left(\frac {x + y}{z}\right)^{r}\geq 3\cdot 2^{r}\] Since $ r>0$ By AM-GM we have \[ \frac {1}{3}\sum \left(\frac {x + y}{z}\right)^{r}\geq \sqrt [3]{\left (\frac {(x + y)(y + z)(x + z)}{xyz}\right )^{r}}\geq 2^{r}\] and because $ r > 0$ the last one is equivalent to \[ (x + y)(y + z)(x + z)\geq 8xyz\]

Svejk wrote: Let $r=-t$ with $t>0$ an the inequality to prove is \[\sum \left ( \frac{x}{y+z}\right)^{t}\geq \frac{3}{2^{t}}\] which has already been posted on this forum. Actually that inequality is not true. For $t=\frac{1}{2}$ for example, the infimum of the LHS is 2. Actually you never get to this case, as $a,b,c>1$ implies that $\ln a, \ln b, \ln c > 0$.

We have $LHS \geq \frac{(log_{a}b+log_{a}c+log_{b}c+log_{b}a+log_{c}a+log_{c}b)^{r}}{3^{r-1}}$ So, we have to prove that $(log_{a}b+log_{a}c+log_{b}c+log_{b}a+log_{c}a+log_{c}b) \geq 6$. See that $log_{a}b+log_{b}a \geq 2\sqrt{log_{a}b.log_{b}a}= 2$, and the problem is solved!

Bonjour Monsieur e.lopez, canst thou please explaineth to me how it is you have so magnificently ascertained that first line of your most delightful proof?

Turkmenistan National Math Olympiad 2012 Grade 9

Here is nice solution. Let $(\log_{a} bc) ^r+(\log_{b} ca) ^r+(\log_{c} ab)^r=X$ From $AM-GM$ inequality we have that $\frac{X}{3} \geq \sqrt[3]{(\log_{a}bc)^r (\log_{b}ca)^r (\log_{c}ab)^r} =[(\log_{a}b+\log_{a}c) (\log_{b}c+\log_{b} a) (\log_{c}a+\log_{c}b)] ^\frac{r} {3}$ and again we can say that this is $\geq [2 \sqrt {\log_{a}b \cdot \log_{a}c} \cdot 2\sqrt {\log_{b}c \cdot \log_{b}a} \cdot 2\sqrt {\log_{c}a \cdot \log_{c}b}] ^\frac{r}{3}$,or $2^r[\log_{a}b \cdot \log_{b}a \cdot \log_{a}c \cdot \log_{c}a \cdot \log_{b}c \cdot \log_{c}b]^\frac{r} {6} =2^r$ So we now get that $X=(\log_{a} bc) ^r+(\log_{b} ca) ^r+(\log_{c} ab) \geq 3 \cdot 2^r$ $Q.E.D$