I am back after a long time, with this question
In a $\Delta ABC; r_1+ r = r_2 + r_3, \angle ABC > \dfrac{\pi}{3}$. Then prove that $b < 3a$.
Here $r_1$ is exradius of excircle formed by internal angle bisector of $\angle BAC$. Similarly other two are defined. $r$ is inradius and $a, b$ are usual notations with respect to a triangle.
I proved it but won't write up the whole solution.
Glimpse of my solution:
I proved $3a > b + 3c$, then manipulated L.H.S. to $3a + 3c > 3a$, which gives the desired answer. But am sure there must be a better way. How should it be done?
Best Answer
In the standard notation $$r+r_a=r_b+r_c$$ gives $$\frac{2S}{a+b+c}+\frac{2S}{b+c-a}=\frac{2S}{a+c-b}+\frac{2S}{a+b-c}$$ or $$a^3+(b+c)a^2-(b+c)^2a-(b+c)(b-c)^2=0.$$ Thus, $$a^3+(b+c)a^2-(b+c)^2a=(b+c)(b-c)^2\geq0,$$ which gives $$a\geq\frac{-b-c+\sqrt{5(b+c)^2}}{2}=\frac{\sqrt5-1}{2}(b+c)>\frac{\sqrt5-1}{2}b>\frac{b}{3}.$$