Prove Inequality: a,b,c>0 \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3

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To prove the inequality \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3\) for positive \(a, b, c\), one effective method is to apply the AM-GM inequality. By clearing the denominators and dividing the expression by 3, the AM-GM inequality can be directly applied to show that the average of the three fractions is at least 1, which leads to the desired result. The discussion highlights the difficulty in manipulating the expression directly and suggests that the inequality holds for all positive real numbers, not just integers. This approach simplifies the proof and confirms the validity of the inequality. The conclusion reinforces that the inequality is universally applicable for positive values of \(a, b,\) and \(c\).
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Homework Statement


If a, b, c > 0, prove \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3


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The Attempt at a Solution


I'm not so sure how to do this. Usually I would try to prove that \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 \ge 0 but this gets me nowhere: \frac{a^2 c + b^2 a + c^2 b - 3abc}{abc}. I can't factorise the numerator.

I know of a similar inequality that I can prove easily using this method, which is \frac{a}{b} + \frac{b}{a} \ge 2 but the inequality above is harder for me. Please help.
 
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Are a, b, and c integers?
 
It doesn't matter if a, b, c are integers or not. The inequality holds for real a,b,c > 0.

Clear denominators, divide by 3, and apply the AM-GM inequality.
 

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