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Algebra Difficulty 4.3 AIME Prove it United States

Problem:

If xx is a positive real number, find the smallest possible value of 2x+18x2x + \frac{18}{x}.

Solution

Solution:

The answer is 1212. This is achieved when x=3x = 3; to see it is optimal, note that
2x+18x12x26x+90, 2x + \frac{18}{x} \geq 12 \Longleftrightarrow x^2 - 6x + 9 \geq 0,
which is obviously true since the left-hand side is (x3)20(x-3)^2 \geq 0.

Alternatively, those who know the so-called AM-GM inequality may apply it directly.

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