Three numbers

By weighted AM-GM inequality, $\sqrt[3]{a^bb^cc^a}\leq\frac{ab+bc+ca}3$

Also by rearrangement inequality we have $a^2+b^2+c^2\geq ab+bc+ca$ .

So $9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca), \frac{ab+bc+ca}3\leq1$

Hence we obtain $a^bb^cc^a\leq \left(\frac{ab+bc+ca}3\right)^3 \leq 1^3=1$

Equality holds when $a=b=c=1$

Let us consider $b$ $a$ 's, $c$ $b$ 's and $a$ $c$ 's. Their A. M. is $\dfrac {ab+bc+ca}{a+b+c}=\dfrac {ab+bc+ca}{3}$

And G. M. is $\sqrt[a+b+c] {a^bb^cc^a}=\sqrt[3] {a^bb^cc^a}$

So, $\sqrt[3] {a^bb^cc^a}\leq \dfrac {ab+bc+ca}{3}$

$\implies a^bb^cc^a\leq (\frac{ab+bc+ca}{3})^3$

Now, $(a-b) ^2+(b-c) ^2+(c-a) ^2\geq 0$

$\implies a^2+b^2+c^2\geq ab+bc+ca$

$\implies (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca)$

$\implies 9\geq 3(ab+bc+ca)\implies ab+bc+ca\leq 3$

So $a^bb^cc^a\leq (\frac{3}{3})^3$ ,

Hence, the maximum value of $a^bb^cc^a$ is $\boxed 1$ .

Consider the value of $\ln (a^bb^cc^a) = a\ln b + b\ln c + c\ln a$ . We note that $\ln(x)$ is a concave function. By Jensen's inequality , we have:

$\begin{aligned} \frac {a\ln b + b\ln c + c\ln a}{a+b+c} \le \ln \left(\frac {ab+bc+ca}{a+b+c} \right) \end{aligned}$

By Hölder's inequality :

$\begin{aligned} (a+b+c)(b+c+a)(1+1+1) & \ge (ab+bc+ca)^3 \\ (3)(3)(3) & \ge (ab+bc+ca)^3 \\ \implies ab+bc+ca & \le 3 \end{aligned}$

Therefore,

$\begin{aligned} \frac {a\ln b + b\ln c + c\ln a}3 & \le \ln \left(\frac 33 \right) = \ln 1 = 0 \\ \implies a\ln b + b\ln c + c\ln a & \le 0 \\ a^b b^c c^a & \le e^0 = \boxed 1 \end{aligned}$