Mathematics - Properties of Triangle Question with Solution | TestHub

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

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

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

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