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

Question

$$\frac{ab+bc+ac}{a^2+b^2+c^2}$$ $$=\frac{\frac 12 ((a+b+c)^2-(a^2+b^2+c^2))}{a^2+b^2+c^2}$$

$$=\frac 12 \left(\frac{(a+b+c)^2}{a^2+b^2+c^2}-1\right)$$

For max value, $a=b=c$

Max =$1$

How do I find the minimum value

Answer 1

In $\Delta ABC$, $a^2=b^2+c^2-2bc \cos A > b^2+c^2-2bc$.

Adding up the similar inequalities gives $ab+bc+ca > \frac{1}{2} (a^2+b^2+c^2)$

Answer 2

Since we have triangle sides, we better use $$a=u+v,\quad b=v+w,\quad c=w+u$$ with $u,v,w>0.$ Arithmetically, $u=(a+c-b)/2, v=(a+b-c)/2, w=(b+c-a)/2,$ so taking $u,v,w>0,$ the triangle inequality is satisfied, automagically. Geometrically speaking, $u,v,w$ are the segments the sides are divided into by the touch-down points of the incircle.

Then, your expression becomes (after some simple algebra) $$\frac{1+3\,p/s}{2\,(1+p/s)},$$ where $s=u^2+v^2+w^2$ and $p=uv+vw+wu.$ Obviously, $0<p\le s,$ so the range is $\left(\frac12,1\right].$