Prove that $a^b\cdot b^c \cdot c^a \leq 1$ I have to prove that

$a^b\cdot b^c\cdot c^a \leq 1$ given that $a+b+c=3$ and $a,b,c$ are positive real numbers

My approach:

By AM-GM,

$a^b\cdot b^c \cdot c^a \leq (\frac{ab+bc+ac}{3})^3$

Now I am not sure how to proceed and in the solution, it goes like this:

${ab+bc+ca\over 3}\leq {a+b+c\over 3}=1$

I dont get the above statement, am I missing something?

Thanks
1=$\frac {(a+b+c)^2}{9}$=$\frac {a^2+b^2+c^2+2ab+2bc+2ca}{9} \ge \frac {ab+bc+ca+2ab+2bc+2ca}{9}$ 

$\Rightarrow 1 \ge \frac {3(ab+bc+ca)}{9}$

$\Rightarrow 1 \ge \frac {(ab+bc+ca)}{3}$

We have used standard inequality that $a^2+b^2+c^2$ $\ge$ ab+bc+ca
By Jensen for the concave function $\ln$ and by the Batominovski's reasoning we obtain:
$$\prod_{cyc}a^b=e^{\sum\limits_{cyc}b\ln{a}}=e^{3\sum\limits_{cyc}\frac{b}{3}\ln{a}}\leq e^{3\ln\left(\sum\limits_{cyc}\left(\frac{b}{3}\cdot a\right)\right)}=e^{3\ln\frac{ab+ac+bc}{3}}\leq e^{3\ln\left(\frac{a+b+c}{3}\right)^2}=1.$$