Triangle Inequalities: Finding the Range of a^2+b^2+c^2/ab+bc+ca

[ritwik06]

Homework Statement

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

The Attempt at a Solution

Let;
$$\frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca}=k$$
cross multiplying and adding 2(ab+bc+ca) on both sides
then;

$$\frac{(a+b+c)^{2}}{ab+bc+ca}=k+2$$

Applying sin rule:
a=2R sin A
b=2R sin B
c=2R sin C
R is the circumradius.
now k+2=$$\frac{(sin A+ sin B+ sin C)^{2}}{sin A sin B+sin B sin C+ sin C sin A}$$
This can be further modified as:
k+2= $$\frac{(sin A+ sin B+ sin C)^{2}}{sin A sin B sin C (cosec A+Cosec B+cosec C)}$$
As A+B+C= 180
I can say that maximum value of $$sin A+sin B+sin C= \frac{3^{1.5}}{2}$$

I also know that the minimum value of $$cosec A+ cosec B+ cosec C=2\sqrt{3}$$
Both these inequalities hold when triangle is equilateral. But I am stuck with the expression sinA. Sin B. sin C? Am I on the right track?

 

[lippyka]

Please help. The range of \frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca} is, \frac{\frac{3^{1.5}}{2}}{2\sqrt{3}} \leq \frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca} \leq \left(\frac{3^{1.5}}{2}\right)^2