Problem
ALG-B2-M07-P018 Product normalization in fractions
#18
★★★★★ Level 5 of 5
Let \(a,b,c>0\). Prove \[\frac{a^2+b^2+c^2}{\sqrt[3]{a^2b^2c^2}}\ge3.\]
Hint. Normalize \(abc=1\).
The expression has degree \(0\). Divide \(a,b,c\) by \(\sqrt[3]{abc}\); the product becomes \(1\), and the expression does not change. With \(abc=1\), we need \(a^2+b^2+c^2\ge3\), which follows from AM-GM.
A more complex form of the same \(abc=1\) normalization.