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

$a^2+b^2+c^2-ab-bc-ca\ge 0$$\Rightarrow \frac{ab+bc+ca}{a^2+b^2+c^2}\le 1$_____________(1) and

$a^2+b^2+c^2=2ab\cos C<2ab$; $b^2+c^2-a^2=2bc\cos A<2bc;c^2+a^2-b^2=2ca\cos B<2ca$

Adding above all,   $(a^2+b^2+c^2)<2(ab+bc+ca)$

$\Rightarrow \frac{1}{2}<\frac{ab+bc+ca}{a^2+b^2+c^2}$______(2)

From (1) and (2)

$\Rightarrow \frac{ab+bc+ca}{a^2+b^2+c^2}\in (\frac{1}{2},1]$