<s>[INST] 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!!! 
 [/INST] You can dress up your geometric argument as an inequality in $\mathbb{C}$.
Let $z=x+iy$, $a=2i$, $b=2-2i$.  Then
\begin{align}
\sqrt{2x^2+2y^2} &= \lvert (1-i)z\rvert\\
\sqrt{y^2+x^2-4y+4} &= \lvert z-a\rvert\\
\sqrt{x^2+y^2-4x-4y+8} &= \lvert -iz-b\rvert
\end{align}
So the triangle inequality gives
$$
\lvert (1-i)z\rvert + \lvert z-a\rvert + \lvert -iz-b\rvert \ge \lvert a-b\rvert
$$
with equality if and only if the four points $a,z,-iz,b$ are all on the same line and in this order.  Now argue why such a $z$ exists.
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