Find the minimum value of $\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$ Given that $0\lt x\lt 2$ and $0\lt y\lt 2$ then find the minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$$
My try: 
On factorisation we need minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {(y-2)^2+x^2} +\sqrt {(x-2)^2+(y-2)^2}$$
On seeing it for first time,  the only thing that popped up was using the Minkowski inequality but I am not getting proper sequences for its application. I tried as much as I could to use this inequality but failed. 
I tried substituting $x=2\cos \alpha$ and $y=2\cos \beta$ (where $\alpha, \beta \in \left(0,\frac {\pi}{2}\right)$ )on seeing the constraints on $x$ and $y$ but continuing it was very cumbersome so dropped the method. 
Any help would be greatly appreciated. 
P.S : It would be very great if someone hints at how I could use the Minkowski inequality efficiently.  Thanks!!! 

You can still use Minkowski as:
$$\sqrt{(x+y)^2+(y-x)^2}+\sqrt{(2-y)^2+x^2}+\sqrt{(2-x)^2+(2-y)^2}\geq\sqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=\sqrt{20}.$$
One can check that the equality is attained when: $$(x,y) = \left(\frac 25,\frac 65\right).$$
