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 , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$

By C-S
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}=2\sqrt{\left(\left(a-\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)}\geq$$
$$2\left(\left(a-\frac{b}{2}\right)\left(x-\frac{y}{2}\right)+\frac{3}{4}by\right)=2ax-bx-ay+2by$$
and we are done!
