I have a triangle ABC and let M be the midpoint of side BC of triangle ABC and let $r_1$ and $r_2$ be the radii of the two incircles of triangles ABM and ACM. I need to prove that $r_1 < 2r_2$.
First of all I feel that the question is quite incomplete since if I interchange $r_1$ and $r_2$ then does it still make complete sense since nothing is mentioned about the lengths of the sides or is it true no matter what I take to be $r_1$ and $r_2$.
However, upon directly using the formula for inradius, and after some gruelling work, if my calculation is correct then I need to prove the following inequality
$$ a^2 + b^2 + 7c^2 - 2ab + 4ac - 8bc < 0$$
How do I prove this?
Let $AM=m.$
Now, $\frac{r_1}{r_2}=\frac{s_1}{s_2}, as \space \Delta_1=\Delta_2.$ (How?)
$$\therefore \frac{r_1}{r_2}=\frac{\frac{a}{2}+m+c}{\frac{a}{2}+m+b}=1+\frac{c-b}{\frac{a}{2}+m+b}$$
Now, try to prove that $$-\frac{1}{2}<\frac{c-b}{\frac{a}{2}+m+b}<1$$
Or you may assume without loss of genarality that $c>b$ and prove just the right hand inequality as both are equivalent(Why?).
Hint: Use triangle inequality.