Problem
ALG-B1-M03-P012 Three Variables and Sum of Squares
#12
★★★☆☆ Level 3 of 5
Real numbers \(x,y,z\) satisfy \(x+y+z=6\), \(x^2+y^2+z^2=12\). Prove that \(xy+yz+zx=12\).
Square \(x+y+z=6\).
\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\).
Substitute the data: \(36=12+2(xy+yz+zx)\). Hence \(xy+yz+zx=12\).
Preparation for symmetric systems in three variables.