Find the minimum value of $sqrt {2x^2+2y^2} +sqrt {y^2+x^2-4y+4} +sqrt {x^2+y^2-4x-4y+8}$












4












$begingroup$


Given that $0lt xlt 2$ and $0lt ylt 2$ 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 $x=2cos alpha$ and $y=2cos beta$ (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!!!






inequality maxima-minima cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 8:01










  • $begingroup$
    @Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
    $endgroup$
    – Digamma
    Dec 8 '18 at 8:06


















4












$begingroup$


Given that $0lt xlt 2$ and $0lt ylt 2$ 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 $x=2cos alpha$ and $y=2cos beta$ (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!!!






inequality maxima-minima cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 8:01










  • $begingroup$
    @Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
    $endgroup$
    – Digamma
    Dec 8 '18 at 8:06
















4












4








4





$begingroup$


Given that $0lt xlt 2$ and $0lt ylt 2$ 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 $x=2cos alpha$ and $y=2cos beta$ (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!!!






inequality maxima-minima cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$




Given that $0lt xlt 2$ and $0lt ylt 2$ 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 $x=2cos alpha$ and $y=2cos beta$ (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!!!






inequality maxima-minima cauchy-schwarz-inequality holder-inequality




inequality maxima-minima cauchy-schwarz-inequality holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 9:03









Martin Sleziak

44.8k9118272




44.8k9118272










asked Dec 8 '18 at 7:54









DigammaDigamma

6,1621440




6,1621440












  • $begingroup$
    Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 8:01










  • $begingroup$
    @Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
    $endgroup$
    – Digamma
    Dec 8 '18 at 8:06




















  • $begingroup$
    Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 8:01










  • $begingroup$
    @Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
    $endgroup$
    – Digamma
    Dec 8 '18 at 8:06


















$begingroup$
Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
$endgroup$
– Rebellos
Dec 8 '18 at 8:01




$begingroup$
Hmm, you seem to hint on some real analysis inequalities. Is this an exercise from such a lesson ?
$endgroup$
– Rebellos
Dec 8 '18 at 8:01












$begingroup$
@Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
$endgroup$
– Digamma
Dec 8 '18 at 8:06






$begingroup$
@Rebellos Maybe, I don't know because it was given to me by one of my friends. I solved it using a geometrical approach but I wanted to know if there could an algebraic solution to it.
$endgroup$
– Digamma
Dec 8 '18 at 8:06












2 Answers
2






active

oldest

votes


















5












$begingroup$

You can still use Minkowski as:
$$sqrt{(x+y)^2+(y-x)^2}+sqrt{(2-y)^2+x^2}+sqrt{(2-x)^2+(2-y)^2}geqsqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=sqrt{20}.$$
One can check that the equality is attained when: $$(x,y) = left(frac 25,frac 65right).$$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    This is what I really wanted. Excellent way to break $x^2+y^2$
    $endgroup$
    – Digamma
    Dec 8 '18 at 9:40



















2












$begingroup$

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)zrvert\
sqrt{y^2+x^2-4y+4} &= lvert z-arvert\
sqrt{x^2+y^2-4x-4y+8} &= lvert -iz-brvert
end{align}
So the triangle inequality gives
$$
lvert (1-i)zrvert + lvert z-arvert + lvert -iz-brvert ge lvert a-brvert
$$
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.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

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    5












    $begingroup$

    You can still use Minkowski as:
    $$sqrt{(x+y)^2+(y-x)^2}+sqrt{(2-y)^2+x^2}+sqrt{(2-x)^2+(2-y)^2}geqsqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=sqrt{20}.$$
    One can check that the equality is attained when: $$(x,y) = left(frac 25,frac 65right).$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      This is what I really wanted. Excellent way to break $x^2+y^2$
      $endgroup$
      – Digamma
      Dec 8 '18 at 9:40
















    5












    $begingroup$

    You can still use Minkowski as:
    $$sqrt{(x+y)^2+(y-x)^2}+sqrt{(2-y)^2+x^2}+sqrt{(2-x)^2+(2-y)^2}geqsqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=sqrt{20}.$$
    One can check that the equality is attained when: $$(x,y) = left(frac 25,frac 65right).$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      This is what I really wanted. Excellent way to break $x^2+y^2$
      $endgroup$
      – Digamma
      Dec 8 '18 at 9:40














    5












    5








    5





    $begingroup$

    You can still use Minkowski as:
    $$sqrt{(x+y)^2+(y-x)^2}+sqrt{(2-y)^2+x^2}+sqrt{(2-x)^2+(2-y)^2}geqsqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=sqrt{20}.$$
    One can check that the equality is attained when: $$(x,y) = left(frac 25,frac 65right).$$






    share|cite|improve this answer









    $endgroup$



    You can still use Minkowski as:
    $$sqrt{(x+y)^2+(y-x)^2}+sqrt{(2-y)^2+x^2}+sqrt{(2-x)^2+(2-y)^2}geqsqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=sqrt{20}.$$
    One can check that the equality is attained when: $$(x,y) = left(frac 25,frac 65right).$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 8 '18 at 9:13









    dezdichadodezdichado

    6,3711929




    6,3711929








    • 2




      $begingroup$
      This is what I really wanted. Excellent way to break $x^2+y^2$
      $endgroup$
      – Digamma
      Dec 8 '18 at 9:40














    • 2




      $begingroup$
      This is what I really wanted. Excellent way to break $x^2+y^2$
      $endgroup$
      – Digamma
      Dec 8 '18 at 9:40








    2




    2




    $begingroup$
    This is what I really wanted. Excellent way to break $x^2+y^2$
    $endgroup$
    – Digamma
    Dec 8 '18 at 9:40




    $begingroup$
    This is what I really wanted. Excellent way to break $x^2+y^2$
    $endgroup$
    – Digamma
    Dec 8 '18 at 9:40











    2












    $begingroup$

    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)zrvert\
    sqrt{y^2+x^2-4y+4} &= lvert z-arvert\
    sqrt{x^2+y^2-4x-4y+8} &= lvert -iz-brvert
    end{align}
    So the triangle inequality gives
    $$
    lvert (1-i)zrvert + lvert z-arvert + lvert -iz-brvert ge lvert a-brvert
    $$
    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.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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)zrvert\
      sqrt{y^2+x^2-4y+4} &= lvert z-arvert\
      sqrt{x^2+y^2-4x-4y+8} &= lvert -iz-brvert
      end{align}
      So the triangle inequality gives
      $$
      lvert (1-i)zrvert + lvert z-arvert + lvert -iz-brvert ge lvert a-brvert
      $$
      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.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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)zrvert\
        sqrt{y^2+x^2-4y+4} &= lvert z-arvert\
        sqrt{x^2+y^2-4x-4y+8} &= lvert -iz-brvert
        end{align}
        So the triangle inequality gives
        $$
        lvert (1-i)zrvert + lvert z-arvert + lvert -iz-brvert ge lvert a-brvert
        $$
        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.






        share|cite|improve this answer











        $endgroup$



        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)zrvert\
        sqrt{y^2+x^2-4y+4} &= lvert z-arvert\
        sqrt{x^2+y^2-4x-4y+8} &= lvert -iz-brvert
        end{align}
        So the triangle inequality gives
        $$
        lvert (1-i)zrvert + lvert z-arvert + lvert -iz-brvert ge lvert a-brvert
        $$
        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '18 at 9:02









        J.G.

        25.9k22540




        25.9k22540










        answered Dec 8 '18 at 8:57









        user10354138user10354138

        7,4322925




        7,4322925






























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