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












2












$begingroup$


Let $x;yin R$. 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:



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



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



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



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.






inequality radicals maxima-minima







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/613028/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 18:22










  • $begingroup$
    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.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 18:32












  • $begingroup$
    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
    $endgroup$
    – Doug M
    Jan 2 at 18:44






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Doug M
    Jan 2 at 18:55








  • 1




    $begingroup$
    Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
    $endgroup$
    – Nguyễn Duy Linh
    Jan 3 at 5:17
















2












$begingroup$


Let $x;yin R$. 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:



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



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



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



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.






inequality radicals maxima-minima







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/613028/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 18:22










  • $begingroup$
    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.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 18:32












  • $begingroup$
    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
    $endgroup$
    – Doug M
    Jan 2 at 18:44






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Doug M
    Jan 2 at 18:55








  • 1




    $begingroup$
    Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
    $endgroup$
    – Nguyễn Duy Linh
    Jan 3 at 5:17














2












2








2


0



$begingroup$


Let $x;yin R$. 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:



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



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



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



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.






inequality radicals maxima-minima







share|cite|improve this question









$endgroup$




Let $x;yin R$. 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:



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



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



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



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.






inequality radicals maxima-minima




inequality radicals maxima-minima






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 18:16









Nguyễn Duy LinhNguyễn Duy Linh

1818




1818












  • $begingroup$
    math.stackexchange.com/questions/613028/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 18:22










  • $begingroup$
    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.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 18:32












  • $begingroup$
    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
    $endgroup$
    – Doug M
    Jan 2 at 18:44






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Doug M
    Jan 2 at 18:55








  • 1




    $begingroup$
    Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
    $endgroup$
    – Nguyễn Duy Linh
    Jan 3 at 5:17


















  • $begingroup$
    math.stackexchange.com/questions/613028/…
    $endgroup$
    – lab bhattacharjee
    Jan 2 at 18:22










  • $begingroup$
    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.
    $endgroup$
    – Michael Rozenberg
    Jan 2 at 18:32












  • $begingroup$
    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
    $endgroup$
    – Doug M
    Jan 2 at 18:44






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Doug M
    Jan 2 at 18:55








  • 1




    $begingroup$
    Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
    $endgroup$
    – Nguyễn Duy Linh
    Jan 3 at 5:17
















$begingroup$
math.stackexchange.com/questions/613028/…
$endgroup$
– lab bhattacharjee
Jan 2 at 18:22




$begingroup$
math.stackexchange.com/questions/613028/…
$endgroup$
– lab bhattacharjee
Jan 2 at 18:22












$begingroup$
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.
$endgroup$
– Michael Rozenberg
Jan 2 at 18:32






$begingroup$
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.
$endgroup$
– Michael Rozenberg
Jan 2 at 18:32














$begingroup$
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
$endgroup$
– Doug M
Jan 2 at 18:44




$begingroup$
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
$endgroup$
– Doug M
Jan 2 at 18:44




2




2




$begingroup$
@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.
$endgroup$
– Doug M
Jan 2 at 18:55






$begingroup$
@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.
$endgroup$
– Doug M
Jan 2 at 18:55






1




1




$begingroup$
Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
$endgroup$
– Nguyễn Duy Linh
Jan 3 at 5:17




$begingroup$
Thx everyone, now i can solve it by Cauchy-Schwarz when i knew equality occurs when $x=y=frac{-1}{sqrt 3}$.
$endgroup$
– Nguyễn Duy Linh
Jan 3 at 5:17










1 Answer
1






active

oldest

votes


















1












$begingroup$

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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
    $endgroup$
    – Martin R
    Jan 2 at 21:14










  • $begingroup$
    @MartinR Thank you. One of the answers there was correct, and was accepted.
    $endgroup$
    – Will Jagy
    Jan 2 at 21:18













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
    $endgroup$
    – Martin R
    Jan 2 at 21:14










  • $begingroup$
    @MartinR Thank you. One of the answers there was correct, and was accepted.
    $endgroup$
    – Will Jagy
    Jan 2 at 21:18


















1












$begingroup$

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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
    $endgroup$
    – Martin R
    Jan 2 at 21:14










  • $begingroup$
    @MartinR Thank you. One of the answers there was correct, and was accepted.
    $endgroup$
    – Will Jagy
    Jan 2 at 21:18
















1












1








1





$begingroup$

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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$



enter image description here






share|cite|improve this answer











$endgroup$



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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 21:23

























answered Jan 2 at 19:43









Will JagyWill Jagy

103k5101200




103k5101200












  • $begingroup$
    It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
    $endgroup$
    – Martin R
    Jan 2 at 21:14










  • $begingroup$
    @MartinR Thank you. One of the answers there was correct, and was accepted.
    $endgroup$
    – Will Jagy
    Jan 2 at 21:18




















  • $begingroup$
    It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
    $endgroup$
    – Martin R
    Jan 2 at 21:14










  • $begingroup$
    @MartinR Thank you. One of the answers there was correct, and was accepted.
    $endgroup$
    – Will Jagy
    Jan 2 at 21:18


















$begingroup$
It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
$endgroup$
– Martin R
Jan 2 at 21:14




$begingroup$
It was closed as a duplicate of math.stackexchange.com/q/2046490/42969 – I cannot tell if those answers are right or wrong, but the result seems to be the same.
$endgroup$
– Martin R
Jan 2 at 21:14












$begingroup$
@MartinR Thank you. One of the answers there was correct, and was accepted.
$endgroup$
– Will Jagy
Jan 2 at 21:18






$begingroup$
@MartinR Thank you. One of the answers there was correct, and was accepted.
$endgroup$
– Will Jagy
Jan 2 at 21:18




















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