# Triangle Inequalities.

1. Oct 15, 2008

### ritwik06

1. The problem statement, all variables and given/known data
a,b,c are the sides of the triangle. then find the range of $$\frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca}$$

3. 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???