Prove that $2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx$? Let $a,b,x,y$ be real numbers , then is it true that 
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx?$$ Actually I am trying to prove the triangle inequality for the norm of numbers in $K(\rho)$ , where $\rho$ is the imaginary cube-root of unity , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$

$4\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)} = \sqrt{(3(a-b)^2+(a+b)^2)(3(x-y)^2+(x+y)^2)} $
$\ge_{c.s.} (3|a-b||x-y|+|a+b||x+y|) \ge 3(a-b)(x-y)+(a+b)(x+y)$
$ = 4ax+4by - 2ay-2bx$
