Question 52
Let $a$, $b$ and $c$ be the sides of a triangle, and we set
$$
A = \frac{a^2 + bc}{b+c} + \frac{b^2 + ca}{c+a} + \frac{c^2 + ab}{a+b}
$$
$$
B = \frac{1}{\sqrt{(a+b-c)(b+c-a)}} + \frac{1}{\sqrt{(b+c-a)(c+a-b)}} + \frac{1}{\sqrt{(c+a-b)(a+b-c)}}
$$
Prove that $AB \ge 9$.
$$
A = \frac{a^2 + bc}{b+c} + \frac{b^2 + ca}{c+a} + \frac{c^2 + ab}{a+b}
$$
$$
B = \frac{1}{\sqrt{(a+b-c)(b+c-a)}} + \frac{1}{\sqrt{(b+c-a)(c+a-b)}} + \frac{1}{\sqrt{(c+a-b)(a+b-c)}}
$$
Prove that $AB \ge 9$.