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

$\begin{array}{rl} & a^2+b^2+c^2-ab-bc-ca\ge 0\\ & a^2+b^2+c^2\ge ab+bc+ca\\ & \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 1\to \left(1\right)\end{array}$
$b^2+c^2-a^2=2bc\cos A<2bc$ Similarly  
$c^2+a^2-b^2<2ca$ and $a^2+b^2-c^2<2ab$ Add above inequalities
$\begin{array}{rl} & \begin{array}{c}{b}^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2\\ <2bc+2ca+2ab\end{array}\\ & a^2+b^2+c^2<2(ab+bc+ca)\\ & \frac{a^2+b^2+c^2}{ab+bc+ca}<2\to \left(2\right)\\ & \text{From (1)\& (2)}\\ & 1\le \frac{a^2+b^2+c^2}{ab+bc+ca}<2\end{array}$