Cunning Maximum

Since $(a-b)^2+(b-c)^2+(c-a)^2\ge0$ , we get: $3(ab+bc+ca)\le(a+b+c)^2=36$ , implying $ab+bc+ca\le 12$

Since $1\le a, b, c\le 3$ , we have:

$\qquad\;\left\{\begin{array}{l}(a-1)(b-1)(c-1)\ge0\\(3-a)(3-b)(3-c)\ge0\end{array}\right.$

Or $\quad\left\{\begin{array}{l}abc-(ab+bc+ca)+5\ge0\\-27+3(ab+bc+ca)-abc\ge0\end{array}\right.$

This implies $2(ab+bc+ca)-22\ge0 \Leftrightarrow ab+bc+ca\ge11$ .

We have:

$P=\dfrac{ (ab)^2 + (bc)^2 + (ca)^2 + 12abc + 72}{ab+bc+ca} - \dfrac{abc}{2}$

$\quad\le\dfrac{(ab)^2+(bc)^2+(ca)^2+2(a+b+c)abc+72}{ab+bc+ca}-\dfrac{ab+bc+ca-5}{2}$

$\quad=\dfrac{(ab+bc+ca)^2+72}{ab+bc+ca}-\dfrac{ab+bc+ca-5}{2}$

$\quad=\dfrac{t^2+72}{t}-\dfrac{t-5}{2}\qquad\qquad$ , where $t=ab+bc+ca$ .

$\quad=\dfrac{t^2+5t+144}{2t}$

$\quad=\dfrac{160}{11}+\dfrac{(t-11)(11t-144)}{22t}$

$\quad\le\dfrac{160}{11}\qquad\qquad$ (since $11\le t\le12$ )

If $a=1, b=2, c=3$ , then $P=\dfrac{160}{11}$

So, the maximum value of $P$ is $\boxed{\dfrac{160}{11}\approx14.545}$