Problem
ALG-B2-M08-P007 Deviations from one
#7
★★★☆☆ Level 3 of 5
Let \(a+b+c=3\). Prove \(a^2+b^2+c^2\ge3\), setting \(a=1+x\), \(b=1+y\), \(c=1+z\).
Hint. Then \(x+y+z=0\).
We have \(x+y+z=0\). Hence \(\sum a^2=\sum(1+x)^2=3+2\sum x+\sum x^2=3+\sum x^2\ge3\).
Teaching goal: recognize the appropriate substitution and check the domain.