Question
If x, y, z are real numbers such that xy+yz+xz = -1, prove x^2+5y^2+8z^2\geq 4
If x, y, z are real numbers such that xy+yz+xz = -1, prove x^2+5y^2+8z^2\geq 4
x^2+5y^2+8z^2 - 4
=x^2+5y^2+8z^2 + 4(xy+yz+xz )
=(x+2y+2z)^2+(y-2z)^2\geq 0
Therefor, x^2+5y^2+8z^2\geq 4