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1+ab+bc+ca >= min { (a + b)^2 / ab } for a,b,c>0 with abc = 1

Source: Switzerland - 2012 Swiss MO Final Round p9

January 14, 2023
inequalitiesalgebra

Problem Statement

Let a,b,c>0a, b, c > 0 be real numbers with abc=1abc = 1. Show 1+ab+bc+ca≥min⁡{(a+b)2ab,(b+c)2bc,(c+a)2ca}.1 + ab + bc + ca \ge \min \left\{ \frac{(a + b)^2}{ab} , \frac{(b+c)^2}{bc} , \frac{(c + a)^2}{ca}\right\}. When does equality holds?
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