Question

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

Collected in the board: Inequality

Steven Zheng posted 1 month ago


Answer

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

Steven Zheng posted 1 month ago

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