Prove that $2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx$?

Question

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

Answer 1

$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$

Answer 2

$$ \begin{align} (u-v)^2&\ge0\tag{1}\\[6pt] u^2v^2+3u^2+3v^2+9&\ge u^2v^2+6uv+9\tag{2}\\[6pt] (u^2+3)(v^2+3)&\ge(uv+3)^2\tag{3}\\[3pt] \sqrt{(2s-1)^2+3}\,\sqrt{(2t-1)^2+3}&\ge\left|\,(2s-1)(2t-1)+3\tag{4}\,\right|\\[3pt] 2\sqrt{s^2-s+1}\,\sqrt{t^2-t+1}&\ge\left|\,2st-s-t+2\,\right|\tag{5}\\ 2\sqrt{\left(\frac ab\right)^2-\frac ab+1}\,\sqrt{\left(\frac xy\right)^2-\frac xy+1}&\ge\left|\,2\frac ab\frac xy-\frac ab-\frac xy+2\,\right|\tag{6}\\ 2\sqrt{a^2-ab+b^2}\,\sqrt{x^2-xy+y^2}&\ge\left|\,2ax-ay-bx+2by\,\right|\tag{7}\\ \end{align} $$ 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}$

Answer 3

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!