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Inequality with condition a+b+c = ab+bc+ca (and special equality case)

Source: BMO 2019, problem 2

May 2, 2019
inequalities

Problem Statement

Let a,b,ca,b,c be real numbers such that 0≤a≤b≤c0 \leq a \leq b \leq c and a+b+c=ab+bc+ca>0.a+b+c=ab+bc+ca >0. Prove that bc(a+1)≥2\sqrt{bc}(a+1) \geq 2 and determine the equality cases.
(Edit: Proposed by sir Leonard Giugiuc, Romania)
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