Problem
ALG-B2-M09-P011 Pairwise sum with fixed product
#11
★★★★☆ Level 4 of 5
Let \(a,b,c>0\), \(abc=1\). Prove \[ab+bc+ca\ge3.\]
Hint. Apply AM-GM to \(ab,bc,ca\).
By AM-GM, \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Equality holds when \(ab=bc=ca=1\), i.e. \(a=b=c=1\).
Fixed product controls \(q\).