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South African Mathematics Olympiad Second Round

Algebra Difficulty 4.9 AIME Prove it South Africa

For two positive real numbers aa and bb, which may be equal, what is the smallest possible value of ab+ba\frac{a}{b} + \frac{b}{a}?

Solution

It is easy to see that if a=ba = b, then the expression is equal to 22, and a few trials will suggest that 22 is the smallest value. A proof is that

Let x=abx = \frac{a}{b}, so x>0x > 0. Then ab+ba=x+1x\frac{a}{b} + \frac{b}{a} = x + \frac{1}{x}.

By the AM-GM inequality:

x+1x≥2x⋅1x=21=2. x + \frac{1}{x} \geq 2\sqrt{x \cdot \frac{1}{x}} = 2\sqrt{1} = 2.

Equality holds when x=1x = 1, i.e., a=ba = b.

Therefore, the smallest possible value is 22.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.