In a Triangle the value a/b+b/c+c/a equal to? Given that..

Question

r_i=Exradius of triangle(i=1,2,3) respectively of the sides a,b,c

Exradius=Intersection of Exterior angle bisectors

Given r₁=2r₂=3r₃

To find value of $$\frac{a}b+\frac{b}c+\frac{c}a$$

My Work till now

$5b=a+5c$
$2a=b+2c$
$3a=3b+c$
by the use of the formula $$r_1=\frac{Area}{s-a}$$ $$s=semiperimeter=\frac{a+b+c}2$$ and similarly for $r_2$and $r_3$ and then canceling area I obtained these relations now unable to calculate further!

Answer 1

By your work we obtain $b=\frac{4}{5}a$ and $c=\frac{3}{5}a$, which gives $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{5}{4}+\frac{4}{3}+\frac{3}{5}=\frac{191}{60}.$$$$

Answer 2

From $$5b=a+5c\\2a=b+2c$$ we get $$10b-10c=2a=b+2c\implies 9b=12c\implies \frac{b}{c}=\frac43$$

Calculate the other fractions similarly.