For real $a$, $b$, $c$ all greater than $1$, show $\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ An inequality question from an olympiad book:

If $a,b,c$ are real numbers satisfying: $a>1$, $b>1$ and $c>1$, then prove that: 
$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$

I tried taking $LCM$, but  it doesn't seem to work.
I don't know how to proceed at all. Just a little hint would be very helpful.
 A: Note that $$0=\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}+\ln\dfrac{c^c}{a^a}=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}$$
Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$
We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$
Thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$
Combining these $3$ inequalities gives us $a\ge b$ and $a\ge c$
Observe that (easy to show) $$\ln\dfrac{a^a}{b^b}\ge \ln\dfrac{a}{b}$$
$$\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}\ge \ln\dfrac{a}{b}+\ln\dfrac{b}{c}$$
Then we have
$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)  \prec  \left(\ln\dfrac{a^a}{b^b}, \ln\dfrac{b^b}{c^c}, \ln\dfrac{c^c}{a^a}\right)$$
or (actually only former majorization holds but showing it requires more work, so, including the possibility is better for the sake of shortness)
$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)  \prec  \left(\ln\dfrac{b^b}{c^c}, \ln\dfrac{a^a}{b^b}, \ln\dfrac{c^c}{a^a}\right)$$
Karamata(Majorization) Inequality with using $f(x)=e^x$ ($f''(x)=e^x>0$) gives the desired inequality
$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
