Maximum maximorum

$abc+ab+bc+ca=(a+1)(b+1)(c+1)-(a+b+c)-1$ let $p=a+1, q=b+1$ and $r=c+1$

$\Rightarrow abc+ab+bc+ca=pqr-13$ and $p+q+r=15$

We know that $\sqrt[3]{pqr}\leq\frac{p+q+r}{3}=\frac{15}{3}=5$ and $\sqrt[3]{pqr}=\frac{p+q+r}{3}$ only is posible by $p=q=r \Leftrightarrow p^3=125$ . Therefore $pqr$ is maximum only if $p=q=r=5$ . So the maximum of $abc+ab+bc+ca=5^3-13=\boxed{112}$

(a+b+c)/3 >= (abc)^1/3 { AM>=GM}

or we get abc<=64 PRODUCT IS MAX WHEN NOS ARE EQUAL so a=4,b=4,c=4

For max/min : i) If the product is given, the sum will be min when the numbers are equal. ii) If sum is given, product will be max when the numbers are equal. Hence, here a+b+c = 12 => a = b = c = 4 Therefore, abc + ab + bc + ca = 4^3 + 3*4^2 = 64 + 48 = 112