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 , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$
 A: $$
\begin{align}
(u-v)^2&\ge0\tag{1}\\[6pt]
u^2v^2+3u^2+3v^2+9&\ge u^2v^2+6uv+9\tag{2}\\[6pt]
(u^2+3)(v^2+3)&\ge(uv+3)^2\tag{3}\\[3pt]
\sqrt{(2s-1)^2+3}\,\sqrt{(2t-1)^2+3}&\ge\left|\,(2s-1)(2t-1)+3\tag{4}\,\right|\\[3pt]
2\sqrt{s^2-s+1}\,\sqrt{t^2-t+1}&\ge\left|\,2st-s-t+2\,\right|\tag{5}\\
2\sqrt{\left(\frac ab\right)^2-\frac ab+1}\,\sqrt{\left(\frac xy\right)^2-\frac xy+1}&\ge\left|\,2\frac ab\frac xy-\frac ab-\frac xy+2\,\right|\tag{6}\\
2\sqrt{a^2-ab+b^2}\,\sqrt{x^2-xy+y^2}&\ge\left|\,2ax-ay-bx+2by\,\right|\tag{7}\\
\end{align}
$$
Explanation:
$\begin{array}{l}
(1):&\text{square of a real is non-negative}\\
(2):&\text{multiply by $3$ and add $u^2v^2+6uv+9$}\\
(3):&\text{factor}\\
(4):&\text{square root, $u=2s-1$ and $v=2t-1$}\\
(5):&\text{expand and divide by $2$}\\
(6):&\text{$s=\frac ab$ and $t=\frac xy$}\\
(7):&\text{multiply by $|by|$}
\end{array}$
A: $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$
A: 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!
