# 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

• The minimal value is not attained, the infimum is $1/2$, if $a=b$ and $c) is small, say. – user436658 Jul 1, 2020 at 11:21 • @ProfessorVector you are right, could you explain the process? Jul 1, 2020 at 11:27 • If those are sides of a triangle,$a,b,c>0$and$a+b>c,\ b+c>a,\ a+c>b$Jul 1, 2020 at 11:35 ## 2 Answers In $$\Delta ABC$$, $$a^2=b^2+c^2-2bc \cos A > b^2+c^2-2bc$$. Adding up the similar inequalities gives $$ab+bc+ca > \frac{1}{2} (a^2+b^2+c^2)$$ • You don't have to use the cosine rule because$a+c>b\implies a>b-c\implies a^2>b^2+c^2-2bc\$, as I mentioned in the comments. It's just a basic triangle inequality. Jul 1, 2020 at 11:41
• Yes it's just a matter of whether triangle inequality or cosine rule is more elementary :) Jul 1, 2020 at 11:43
• Well, I heard of the triangle inequality in the elementary school, whereas trigonometry is taught in highschool (in my country). (: Jul 1, 2020 at 11:44

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, $$0 so the range is $$\left(\frac12,1\right].$$

• You could've use the triangle inequality to find the infimum. (: Jul 1, 2020 at 11:49