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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.

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$$ \overline{AB} = \sqrt{(x+u)^{2} + (x+u)(y+v) + (y+v)^{2}} \\ \overline{BC} = \sqrt{x^{2} + xy + y^{2}} \\ \overline{CA} = \sqrt{u^{2} + uv + v^{2}} $$

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    $\begingroup$ Very nice use of the triangle! +1 $\endgroup$ – Macavity Apr 14 at 18:23
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    $\begingroup$ Nice...........+1 $\endgroup$ – Maria Mazur Apr 14 at 18:38
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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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