Library / /8 of 16

, 2019

Algebra Difficulty 8.1 Shortlist Prove it Turkey

Find the minimal possible value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ over all positive real numbers $a, b, c$ satisfying
$abc = 1, \quad a+b+c = 5 \text{ and}$
$(ab+2a+2b-9)(bc+2b+2c-9)(ca+2c+2a-9) \geq 0.$

Solution

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, $(abc = 1)$
$(ab + 2a + 2b - 9)(bc + 2b + 2c - 9)(ca + 2c + 2a - 9)$
---

$= (2a+1)(2b+1)(2c+1)(1-a)(1-b)(1-c) \geq 0.$
$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}).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.