If a, b, c are the sides of a triangle then the range of  $\frac{a^2+b^2+c^2}{ab+bc+ca}$  is

$\text{W.K.T\hspace{0.17em}}{a}^2+b^2+c^2-ab-bc-ca\ge 0\Rightarrow a^2+b^2+c^2\ge ab+bc+ca$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 1$

From  $b^2+c^2-a^2=2bc\cos A<2bc$ rule  $\begin{array}{l}{c}^2+a^2-b^2=2ca\cos B<2ca\\ a^2+b^2-c^2=2ba\cos C<2ab\end{array}$

$a^2+b^2+c^2<2(ab+bc+ca)\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2$

$\therefore 1\le \frac{a^2+b^2+c^2}{ab+bc+ca}<2$