Problem
ALG-B2-M07-P004 Product normalization
#4
★★☆☆☆ Level 2 of 5
Prove for \(a,b,c>0\): \[a+b+c\ge3\sqrt[3]{abc}.\]
Hint. Divide all variables by \(\sqrt[3]{abc}\).
Set \(x=a/\sqrt[3]{abc}\), \(y=b/\sqrt[3]{abc}\), \(z=c/\sqrt[3]{abc}\). Then \(xyz=1\). We need \(x+y+z\ge3\), which follows from AM-GM.
Shows when the normalization \(abc=1\) is appropriate.