Ta có : Q= (x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)
=x² +y² +z²+ 3xy +3yz+3xz
P= (x+y)²+(y+z)²+(z+x)²
= x²+2xy+y²+y²+2yz+z²+z²+2zx+x²
=2x²+2y²+2z²+2xy+2yz+2xz
Lại có P - Q = 2x² + 2y² + 2z² + 2xy + 2yz + 2xz - x² - y² -z²- 3xy -3yz-3xz
= x²+y²+z² - xy - yz - xz=0
=> x²+y²+z² = xy + yz +xz
mà x²+y²+z² $\geq$ xy + yz +xz
=> dấu bằng xảy ra khi x = y = z khi P=Q