<s>[INST] If $a,b,c$ are sides of a triangle, then find range of $\frac{ab+bc+ac}{a^2+b^2+c^2}$ $$\frac{ab+bc+ac}{a^2+b^2+c^2}$$
$$=\frac{\frac 12 ((a+b+c)^2-(a^2+b^2+c^2))}{a^2+b^2+c^2}$$
$$=\frac 12 \left(\frac{(a+b+c)^2}{a^2+b^2+c^2}-1\right)$$
For max value, $a=b=c$
Max =$1$
How do I find the minimum value
 [/INST] In $\Delta ABC$, $a^2=b^2+c^2-2bc \cos A > b^2+c^2-2bc$.
Adding up the similar inequalities gives $ab+bc+ca > \frac{1}{2} (a^2+b^2+c^2)$
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