Problem - 2461
Let $a, b, c$ be respectively the lengths of three sides of a triangle, and $r$ be the triangle's inradius. Show that $$r = \frac{1}{2}\sqrt{\frac{(b+c-a)(c+a-b)(b+a-c)}{a+b+c}}$$
Let $S$ be the area of $\triangle{ABC}$. By the Heron's formula, we have $$S=\sqrt{p(p-a)(p-b)(p-c)}$$ where $p =\frac{a+b+c}{2}$. Therefore, we have $$S=\frac{1}{4}\cdot\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$$
Meanwhile, we also have $$S=r\cdot p = r \cdot \frac{a+b+c}{2}$$
Setting these two equation equal will yield the desired result immediately.