Team Selection Test for EGMO 2019

Answer: 5. Since $abc=1$ we find the minimal value of $ab+bc+ac=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Note that $$ab+2a+2b+2c-9=\frac{1}{c}+2(5-c)-9=\frac{1}{c}-2c+1=\frac{1}{c}(2c+1)(1-c).$$ The similar formulas are held for $bc+2b+2c-9$ and $ca+2c+2a-9$. Therefore, $(abc=1)$ $$(ab+2a+2b-9)(bc+2b+2c-9)(ca+2c+2a-9)$$ ---

$=(2a+1)(2b+1)(2c+1)(1-a)(1-b)(1-c)\ge 0.$ Now since $(2a+1)(2b+1)(2c+1) > 0$, we get 0 \le (1-a)(1-b)(1-c) = -abc - a - b - c + ab + bc + ac + 1. Thus, $ab + bc + ac \ge 5$. The equality holds at (a, b, c) = (1, 2 - \sqrt{3}, 2 + \sqrt{3}). $$