Problem
ALG-B2-M09-P016 Product of pairwise sums
#16
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[(a+b)(b+c)(c+a)\le8.\]
Hint. Notice that \(a+b=3-c\), and apply AM-GM to \(3-a,3-b,3-c\).
We have \((a+b)(b+c)(c+a)=(3-c)(3-a)(3-b)\). The sum of the factors is \((3-a)+(3-b)+(3-c)=6\). By AM-GM, their product is at most \((6/3)^3=8\). Equality holds at \(a=b=c=1\).
A less obvious constraint becomes a fixed sum of new factors.