Minimum value of $sqrt {(x+2)^2+(y+2)^2}+sqrt {(x+1)^2+(y-1)^2}+sqrt{ (x-1)^2+(y+1)^2}$
$begingroup$
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
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
$endgroup$
|
show 3 more comments
$begingroup$
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
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
$endgroup$
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
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
asked Jan 2 at 18:16
Nguyễn Duy LinhNguyễn Duy Linh
1818
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
$endgroup$
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
$$ sqrt 6 + sqrt 8 approx 5.277916867529368195800661523 $$
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
edited Jan 2 at 21:23
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
answered Jan 2 at 19:43
Will JagyWill Jagy
103k5101200
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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