22nd Korean Mathematical Olympiad Final Round

Clearly $$B\ge \frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$ and $$A-(a+b+c)=\frac{a^4+b^4+c^4-a^2{b}^2-b^2{c}^2-c^2{a}^2}{(a+b)(b+c)(c+a)}\ge 0.$$ Therefore we have $$AB\ge \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)(b+c+a)\ge 9.$$ by Cauchy-Schwarz inequality. This completes the proof. □