$a+OA\lt b+OB\lt c+OC$ when $a\lt b\lt c$ in a triangle

Question

In the triangle $\triangle ABC$ of sides $a,b,c$ let $O$ be the incenter. If $a\lt b\lt c$ then (it is easy to prove that) $OC\lt OB\lt OA$. Prove that $$\max \{a+OA, b+OB, c+OC\}=c+OC$$

Answer 1

We need to prove that $$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>b+\sqrt{\left(\frac{a+c-b}{2}\right)^2+r^2}$$ and$$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>a+\sqrt{\left(\frac{b+c-a}{2}\right)^2+r^2}.$$ We'll prove a first inequality. The second inequality can be proved by the same way.

We need to prove that $$c-b>\frac{1}{2}\left(\sqrt{(a+c-b)^2+4r^2}-\sqrt{(a+b-c)^2+4r^2}\right)$$ or $$2(c-b)>\frac{(a+c-b)^2-(a+b-c)^2}{\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}}$$ or $$\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}>2a,$$ which is true because $$\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}>a+c-b+a+b-c=2a.$$