Let D be the inscribed circle of a triangle. There must be three other lesser inscribed circles, each touching D and two sides of the triangle.
I want to show that
$$r \leq r_a+r_b+r_c$$
All I can see is three isosceles triangles to the vertices though. Is there an inequality regarding inscribed circles that I can make use of here?
Any suggestion will be helpful. Thanks.
Let $\alpha$, $\beta$ and $\gamma$ be the angles of the vertices $A$, $B$ and $C$, respectively. From the similar triangles seen in the diagram, we have
$$\frac{r_c}{r}=\frac{CD_c}{CD}=\frac{CD_c}{CD_c +r_c+r} $$
Use $CD_c=r_c/\sin(\gamma/2)$ to reexpress above equation in the form of ratio,
$$\frac{r_c}{r} = \frac{1-\sin\frac{\gamma}{2}}{1+\sin\frac{\gamma}{2}} =\cot^2\left(\frac{\pi+\gamma}{4}\right) $$
Similar expressions can be derived for the other two ratios. Together, we have
$$\frac{r_a+r_b+r_c}{r}= \cot^2\left(\frac{\pi+\alpha}{4}\right)+\cot^2\left(\frac{\pi+\beta}{4}\right)+\cot^2\left(\frac{\pi+\gamma}{4}\right)$$
Use the fact that, for $0<\theta<\pi/2$, $\cot^2(\theta)$ is a convex function to establish the inequality below,
$$\cot^2\left(\frac{\pi+\alpha}{4}\right)+\cot^2\left(\frac{\pi+\beta}{4}\right)+\cot^2\left(\frac{\pi+\gamma}{4}\right)$$ $$ \ge 3\cot^2\left( \frac{\pi}{4} + \frac{\alpha+\beta+\gamma}{12} \right) = 3\cot^2 \left( \frac{\pi}{3} \right) = 1$$
In the end, we have, $$r_a+r_b+r_c \ge r$$