$\text{ If }a,b,c\text{ are sides of a triangle satisfying }{a}^2+b^2+c^2=6,\text{ then the }AM\text{ of all the }$$\text{ integral values which lie in the range of }ab+bc+ca\text{ is }$

$\frac{1}{2}\left\{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right\}\ge 0\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\Rightarrow 6\ge ab+bc+ca$

$a^2=b^2+c^2-2bc\cos A>b^2+c^2-2bc \left(\because \cos A<1\right)$

$b^2>a^2+c^2-2ac\text{ and }{c}^2>a^2+b^2-2ab$

$\therefore {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2>2\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)$

$\Rightarrow ab+bc+ca>\frac{1}{2}\left(a^2+b^2+c^2\right)=3.\text{ hence }ab+bc+ca\in (3,6]$