Minimum value of $\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$

Question

Let $x;y\in R$. Find Minimum value of the function $$\sqrt {(x+2)^2+(y+2)^2}+\sqrt {(x+1)^2+(y-1)^2}+\sqrt{ (x-1)^2+(y+1)^2}$$

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My try: By Minkowski inequality:

$LHS=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+2\sqrt{\dfrac{1}{4}[(x+2)^2+(y+2)^2]}$

$=\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x-1)^2+(y+1)^2}+2\sqrt{(\dfrac{x}{2}+1)^2+(\dfrac{y} {2}+1)^2]}$

$\ge \sqrt{(\dfrac{3x}{2}+2)^2+(\dfrac{3y}{2})^2}+\sqrt {(\dfrac{3y}{2}+2)^2+(\dfrac{3x}{2})^2}\ge \sqrt{8}$

And the equality occurs when $x=-y-\frac {4}{3}$

Help me check it. I fear that is not minimum value of the function. THx.

Answer 1

FERMAT POINT

I get, from the $120^\circ$ construction, that the minimizing point is at $$ \left(\frac{-1}{\sqrt 3},\frac{-1}{\sqrt 3} \right)$$ The main calculation I did was $45^\circ + 60^\circ = 105^\circ$ and $\tan 105^\circ = -2-\sqrt 3.$ The sum of three distances is $$ \sqrt 6 + \sqrt 8 \approx 5.277916867529368195800661523 $$