TITLE: $\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$
QUESTION [2 upvotes]: Let $x,y,u,v \in \mathbb{R}.$ Prove that
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$

Proof 1:
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Square both side, we have
$$2\sqrt{(x^2+xy+y^2)(u^2+uv+v^2)} \geq 2xu+xv+yu+2yv$$
If $2xu+xv+yu+2yv<0$ then the inequality is true.
If $2xu+xv+yu+2yv \geq 0$ the square both side, we have
$$x^2v^2-2xvyu+y^2u^2 \geq 0$$
$$ \Leftrightarrow (xv-yu)^2 \geq 0$$
which is true.
Thus the inequality is true for all $x,y,u,v \in \mathbb{R}.$

Proof 2 (need help):
Because
$$x^2+xy+y^2 \geq \frac{3}{4}(x+y)^2$$
for all $x,y \in \mathbb{R},$ we have
$$\begin{matrix}
\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} & \geq & \frac{\sqrt{3}}{2}(|x+y|+|u+v|)\\ 
 & \geq & \frac{\sqrt{3}}{2}|x+u+y+v|\\ 
 & = & \frac{\sqrt{3}}{2}\sqrt{(x+u+y+v)^2}\\ 
 & = & \frac{\sqrt{3}}{2}\sqrt{(x+u)^2+2(x+u)(y+v)+(y+v)^2}
\end{matrix}$$
I'm stuck here. Do you have any idea? Thank you.

REPLY [1 votes]: By Minkowski (the triangle inequality) we obtain:
$$\sqrt{x^2+xy+y^2}+\sqrt{u^2+uv+v^2}=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2}+\sqrt{\left(u+\frac{v}{2}\right)^2+\frac{3}{4}v^2}\geq$$
$$\geq\sqrt{\left(x+u+\frac{y+v}{2}\right)^2+\frac{3}{4}(y+v)^2}=\sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}.$$
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