<s>[INST] $a+OA\lt b+OB\lt c+OC$ when $a\lt b\lt c$ in a triangle 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
$$\max \{a+OA, b+OB, c+OC\}=c+OC$$
 [/INST] 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
$$\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}>a+c-b+a+b-c=2a.$$
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