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

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.

nDLynk
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  • https://math.stackexchange.com/questions/613028/find-the-point-which-has-shortest-sum-of-distance-from-all-points – lab bhattacharjee Jan 02 '19 at 18:22
  • In my opinion it's not duplicate. The trying to find the solution of Nguyễn Duy Linh is much more better than all linked solutions. – Michael Rozenberg Jan 02 '19 at 18:32
  • This function is symmetric. i.e. if you swap the x's and the y's you have the same function. That suggests that line $x=y$ is special – Doug M Jan 02 '19 at 18:44
  • @Doug Are you sure? We can think about this during construction of the proof, but we can not use it. – Michael Rozenberg Jan 02 '19 at 18:48
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    @MichaelRozenberg The function is symmetric. But that doesn't guarantee that that minimum lies on the line $x= y$ It could be that there is a minimum on one side and its pair on the other, with a saddle on the line $x=y$ Nonetheless, it gives a good place to begin the investigation. – Doug M Jan 02 '19 at 18:55
  • I agree with this statement. But we can not assume $x=y$. – Michael Rozenberg Jan 02 '19 at 18:57
  • I opened this topic because Nguyễn Duy Linh look for a solution by Minkowski and we need to help. I think for three points it's possible. – Michael Rozenberg Jan 02 '19 at 20:21
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    Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=\frac{-1}{\sqrt 3}$. – nDLynk Jan 03 '19 at 05:17

1 Answers1

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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 $$ \sqrt 6 + \sqrt 8 \approx 5.277916867529368195800661523 $$

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Will Jagy
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  • It was closed as a duplicate of https://math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same. – Martin R Jan 02 '19 at 21:14
  • @MartinR Thank you. One of the answers there was correct, and was accepted. – Will Jagy Jan 02 '19 at 21:18