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Exercise 1739Algebra › InequalitiesOlympiadAdded 3 Nov 2026

A ratio-sum inequality

#ratios#AM-GM#cyclic sums

Statement

Let a,b,c>0a,b,c>0. Prove that
ab+bc+ca≥3.\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3.

Hint

Use AM-GM on the three terms.

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Solution

AM-GM.
ab+bc+ca≥3ab⋅bc⋅ca3=313=3.\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=3\sqrt[3]{1}=3.

Equality case. ab=bc=ca\frac ab=\frac bc=\frac ca, which implies a=b=ca=b=c.

Conclusion. The inequality holds, with equality iff a=b=ca=b=c.