Problem
ALG-B2-M09-P018 Squares with fixed product
#18
★★★★★ Level 5 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, \(\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}=1\). Thus \(a^2+b^2+c^2\ge3\). Equality holds at \(a=b=c=1\).
Fixed product for a power sum.