Problem
ALG-B2-M07-P012 Squares with fixed product
#12
★★★★☆ Level 4 of 5
Let \(a,b,c>0\), \(abc=1\). Prove \[a^2+b^2+c^2\ge3.\]
Hint. Apply AM-GM to \(a^2,b^2,c^2\).
By AM-GM, \(\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}=1\). Therefore \(\sum a^2\ge3\).
A simple \(abc=1\) problem, useful for later substitutions.