Minimum value of $P=\sqrt{2x^2+2y^2-2x+2y+1}+\sqrt{2x^2+2y^2+2x-2y+1}+\sqrt{2x^2+2y^2+4x+4y+4}$ Let $x,y∈R$ . Find the minimum value of this expression: 
$P=\sqrt{2x^2+2y^2-2x+2y+1}+\sqrt{2x^2+2y^2+2x-2y+1}+\sqrt{2x^2+2y^2+4x+4y+4}$
We have: $P=\sqrt{(\sqrt{2}x-\frac{1}{\sqrt{2}})^2+(\sqrt{2}y+\frac{1}{\sqrt{2}})^2}+...+\sqrt{(\sqrt{2}x+\sqrt{2})^2+(\sqrt{2}y+\sqrt{2})^2}$
I think use vector or Geometric method to solve this problem. But I don't know how to choose vectors or points logically?

Hint:
You need $P(x, y)$ to be the Fermat point of the triangle $A(\frac12, -\frac12), B(-\frac12, \frac12), C(-1, -1)$. 
Note that $\triangle ABC$ is isosceles, and from properties of the Fermat Point, it should lie on the angle bisector at $C$.  This gives us $x=y$. Now there are two good routes,
(i) to use calculus in one variable, or
(ii) to note an additional property of the Fermat point $\angle APB = \angle BPC = \angle CPA = 120^{\circ}$ and use coordinate geometry / vectors.
Added:  To check your final answer, 

 I get the minimum as $P \ge 2+\sqrt3$.

