Problem
ALG-B1-M01-P019 Symmetric Triple
#19
★★★★☆ Level 4 of 5
Real numbers \(x,y,z\) satisfy \(x+y+z=3\) and \(x^2+y^2+z^2=3\). Prove that \(x=y=z=1\).
Consider \((x-1)^2+(y-1)^2+(z-1)^2\).
Compute: \[ (x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3. \] By the conditions this is \(3-6+3=0\).
A sum of squares is zero only when each square is zero. Hence \(x=y=z=1\).
The problem teaches building the right sum of squares from the conditions.