Problem
ALG-B1-M11-P017 Equality case
#17
★★★★☆ Level 4 of 5
Let \(a,b,c>0\), \(a+b+c=3\). Prove \(a^2+b^2+c^2\ge3\).
Equality is expected at \(a=b=c=1\).
By Cauchy, \((a^2+b^2+c^2)(1+1+1)\ge(a+b+c)^2=9\). Hence \(a^2+b^2+c^2\ge3\).
Strategy: the equality case suggests uniformity.