Question

Đã trả lời bởi chuyên gia

Answer 1

Đặt $x = b + c - a$

$y = a + c - b$

$z = a + b - c$

$\Rightarrow x + z = b + c - a + a + b - c = 2 b$

$x + y = b + c - a + a + c - b = 2 c$

$y + z = a + c - b + a + b - c = 2 a$

Ta có: $2A = \frac{y + z}{x} + \frac{x + z}{y} + \frac{x + y}{z}$

$= \frac{y}{x} + \frac{z}{x} + \frac{x}{y} + \frac{z}{y} + \frac{x}{z} + \frac{y}{z}$

$= (\frac{y}{x} + \frac{x}{y}) + (\frac{z}{y} + \frac{y}{z}) + (\frac{x}{z} + \frac{z}{x})$

Dễ chứng minh $\frac{a}{b} + \frac{b}{a} \geq 2$ với a, b > 0

$\Rightarrow 2A = (\frac{y}{x} + \frac{x}{y}) + (\frac{z}{y} + \frac{y}{z}) + (\frac{x}{z} + \frac{z}{x}) \geq 6$

$\Rightarrow 2A = \frac{y + z}{x} + \frac{x + z}{y} + \frac{x + y}{z} \geq 6$

$\Rightarrow 2A = \frac{2a}{b + c - a} + \frac{2b}{a + c - b} + \frac{2c}{a + b - c} \geq 6$

$\Rightarrow A = \frac{a}{b + c - a} + \frac{b}{a + c - b} + \frac{c}{a + b - c} \geq 3$ (Đpcm)

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Answer 2