Cho a, b dương thoả mãn a - b = a / b. Chứng minh rằng: (ab/(a + b)) × (1/(a + b) + 1/(ab - a - b)) + 1/(ab - a - b) >= 9/ab

Bài toán yêu cầu chứng minh bất đẳng thức:

\[\frac{ab}{a+b} \cdot \left(\frac{1}{a+b} + \frac{1}{ab-a-b}\right) + \frac{1}{ab-a-b} \geq \frac{9}{ab}\]

Đặt \(x = a+b\), ta có \(a = x - b\).

Thay \(a = x - b\) vào \(a - b = \frac{a}{b}\), ta được:

\(x - 2b = \frac{x - b}{b}\)

\(x - 2b = \frac{x}{b} - 1\)

\(x = \frac{x}{b} + b - 1\)

\(x = \frac{x + b^2 - b}{b}\)

\(x = \frac{x + b^2 - b}{b}\)

\(x = \frac{x + b(b - 1)}{b}\)

\(x = \frac{x}{b} + b - 1\)

\(x = \frac{x}{b} + b - 1\)

\(x = \frac{x}{b} + b - 1\)

\(x = \frac{x}{b} + b - 1\)

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