<s>[INST] Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$ 
If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$

Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.
 [/INST] We know from triangle inequality 
In a triangle the sum of any two sides is greater  than the third side, from which it follows that 
$(a+b-c)(a+c-b)+(b+c-a)(b+a-c)+(c+a-b)(c+b-a)>0$
Just expand this to get the inequality 
$\Rightarrow a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ac+c^2-a^2-b^2+2ab>0$
$\Rightarrow -a^2-b^2-c^2+2(ab+bc+ca)>0$
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