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

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

For max value, $a=b=c$

Max =$1$

How do I find the minimum value

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    $\begingroup$ The minimal value is not attained, the infimum is $1/2$, if $a=b$ and $c) is small, say. $\endgroup$
    – user436658
    Jul 1, 2020 at 11:21
  • $\begingroup$ @ProfessorVector you are right, could you explain the process? $\endgroup$
    – Aditya
    Jul 1, 2020 at 11:27
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    $\begingroup$ If those are sides of a triangle, $a,b,c>0$ and $a+b>c,\ b+c>a,\ a+c>b$ $\endgroup$
    – PinkyWay
    Jul 1, 2020 at 11:35

2 Answers 2

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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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    $\begingroup$ 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. $\endgroup$
    – PinkyWay
    Jul 1, 2020 at 11:41
  • $\begingroup$ Yes it's just a matter of whether triangle inequality or cosine rule is more elementary :) $\endgroup$
    – user600016
    Jul 1, 2020 at 11:43
  • $\begingroup$ Well, I heard of the triangle inequality in the elementary school, whereas trigonometry is taught in highschool (in my country). (: $\endgroup$
    – PinkyWay
    Jul 1, 2020 at 11:44
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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 $\left(\frac12,1\right].$

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    $\begingroup$ You could've use the triangle inequality to find the infimum. (: $\endgroup$
    – PinkyWay
    Jul 1, 2020 at 11:49

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