How to show $x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4) > 0$












3












$begingroup$


I have the following information. All are labels/variables.



$x_1 < x_3 < x_4 < x_2$



$y_1 < y_3 < y_4 < y_2$



$x_1 + x_2 = x_3 + x_4$



$x_1 + a = x_3$



$x_2 - a = x_4$




  • a > 0

  • all x and y values are postive


Is there any possibility for me to show that
$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4) > 0$ ?



What I've so far is



$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4)$



$x_1y_1 + x_2y_2 - ((x_1 + a)y_3 + (x_2 - a)y_4)$



$x_2(y_2 - y_4) + a(y_4 - y_3) - x_1(y_3 - y_1)$



In the above statement, the first 2 parts are positive. But I'm stuck to show that the difference between the first 2 parts and the third part is positive






linear-algebra proof-verification proof-writing







share|cite|improve this question











$endgroup$



migrated from mathematica.stackexchange.com Jan 17 at 11:23


This question came from our site for users of Wolfram Mathematica.


















  • $begingroup$
    What do "x1" "y1" etc. mean? Does it mean $x^1$?
    $endgroup$
    – Puffy
    Jan 17 at 11:29










  • $begingroup$
    yeah those are just label names :)
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:30










  • $begingroup$
    are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
    $endgroup$
    – Milan Stojanovic
    Jan 17 at 11:32










  • $begingroup$
    @MilanStojanovic those are just variables/labels
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:33










  • $begingroup$
    $x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
    $endgroup$
    – user376343
    Jan 17 at 11:37
















3












$begingroup$


I have the following information. All are labels/variables.



$x_1 < x_3 < x_4 < x_2$



$y_1 < y_3 < y_4 < y_2$



$x_1 + x_2 = x_3 + x_4$



$x_1 + a = x_3$



$x_2 - a = x_4$




  • a > 0

  • all x and y values are postive


Is there any possibility for me to show that
$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4) > 0$ ?



What I've so far is



$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4)$



$x_1y_1 + x_2y_2 - ((x_1 + a)y_3 + (x_2 - a)y_4)$



$x_2(y_2 - y_4) + a(y_4 - y_3) - x_1(y_3 - y_1)$



In the above statement, the first 2 parts are positive. But I'm stuck to show that the difference between the first 2 parts and the third part is positive






linear-algebra proof-verification proof-writing







share|cite|improve this question











$endgroup$



migrated from mathematica.stackexchange.com Jan 17 at 11:23


This question came from our site for users of Wolfram Mathematica.


















  • $begingroup$
    What do "x1" "y1" etc. mean? Does it mean $x^1$?
    $endgroup$
    – Puffy
    Jan 17 at 11:29










  • $begingroup$
    yeah those are just label names :)
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:30










  • $begingroup$
    are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
    $endgroup$
    – Milan Stojanovic
    Jan 17 at 11:32










  • $begingroup$
    @MilanStojanovic those are just variables/labels
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:33










  • $begingroup$
    $x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
    $endgroup$
    – user376343
    Jan 17 at 11:37














3












3








3





$begingroup$


I have the following information. All are labels/variables.



$x_1 < x_3 < x_4 < x_2$



$y_1 < y_3 < y_4 < y_2$



$x_1 + x_2 = x_3 + x_4$



$x_1 + a = x_3$



$x_2 - a = x_4$




  • a > 0

  • all x and y values are postive


Is there any possibility for me to show that
$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4) > 0$ ?



What I've so far is



$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4)$



$x_1y_1 + x_2y_2 - ((x_1 + a)y_3 + (x_2 - a)y_4)$



$x_2(y_2 - y_4) + a(y_4 - y_3) - x_1(y_3 - y_1)$



In the above statement, the first 2 parts are positive. But I'm stuck to show that the difference between the first 2 parts and the third part is positive






linear-algebra proof-verification proof-writing







share|cite|improve this question











$endgroup$




I have the following information. All are labels/variables.



$x_1 < x_3 < x_4 < x_2$



$y_1 < y_3 < y_4 < y_2$



$x_1 + x_2 = x_3 + x_4$



$x_1 + a = x_3$



$x_2 - a = x_4$




  • a > 0

  • all x and y values are postive


Is there any possibility for me to show that
$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4) > 0$ ?



What I've so far is



$x_1y_1 + x_2y_2 - (x_3y_3 + x_4y_4)$



$x_1y_1 + x_2y_2 - ((x_1 + a)y_3 + (x_2 - a)y_4)$



$x_2(y_2 - y_4) + a(y_4 - y_3) - x_1(y_3 - y_1)$



In the above statement, the first 2 parts are positive. But I'm stuck to show that the difference between the first 2 parts and the third part is positive






linear-algebra proof-verification proof-writing




linear-algebra proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 12:18







Thusitha Thilina Dayaratne

















asked Jan 17 at 11:21









Thusitha Thilina DayaratneThusitha Thilina Dayaratne

1163




1163




migrated from mathematica.stackexchange.com Jan 17 at 11:23


This question came from our site for users of Wolfram Mathematica.









migrated from mathematica.stackexchange.com Jan 17 at 11:23


This question came from our site for users of Wolfram Mathematica.














  • $begingroup$
    What do "x1" "y1" etc. mean? Does it mean $x^1$?
    $endgroup$
    – Puffy
    Jan 17 at 11:29










  • $begingroup$
    yeah those are just label names :)
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:30










  • $begingroup$
    are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
    $endgroup$
    – Milan Stojanovic
    Jan 17 at 11:32










  • $begingroup$
    @MilanStojanovic those are just variables/labels
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:33










  • $begingroup$
    $x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
    $endgroup$
    – user376343
    Jan 17 at 11:37


















  • $begingroup$
    What do "x1" "y1" etc. mean? Does it mean $x^1$?
    $endgroup$
    – Puffy
    Jan 17 at 11:29










  • $begingroup$
    yeah those are just label names :)
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:30










  • $begingroup$
    are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
    $endgroup$
    – Milan Stojanovic
    Jan 17 at 11:32










  • $begingroup$
    @MilanStojanovic those are just variables/labels
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 11:33










  • $begingroup$
    $x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
    $endgroup$
    – user376343
    Jan 17 at 11:37
















$begingroup$
What do "x1" "y1" etc. mean? Does it mean $x^1$?
$endgroup$
– Puffy
Jan 17 at 11:29




$begingroup$
What do "x1" "y1" etc. mean? Does it mean $x^1$?
$endgroup$
– Puffy
Jan 17 at 11:29












$begingroup$
yeah those are just label names :)
$endgroup$
– Thusitha Thilina Dayaratne
Jan 17 at 11:30




$begingroup$
yeah those are just label names :)
$endgroup$
– Thusitha Thilina Dayaratne
Jan 17 at 11:30












$begingroup$
are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
$endgroup$
– Milan Stojanovic
Jan 17 at 11:32




$begingroup$
are x1,x2,x3,x4 4 variables or one variable raised to powers. question by puffy confused me
$endgroup$
– Milan Stojanovic
Jan 17 at 11:32












$begingroup$
@MilanStojanovic those are just variables/labels
$endgroup$
– Thusitha Thilina Dayaratne
Jan 17 at 11:33




$begingroup$
@MilanStojanovic those are just variables/labels
$endgroup$
– Thusitha Thilina Dayaratne
Jan 17 at 11:33












$begingroup$
$x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
$endgroup$
– user376343
Jan 17 at 11:37




$begingroup$
$x1$ is $x_1$ and so on, yes? Are all $x$ and $y$ positive?
$endgroup$
– user376343
Jan 17 at 11:37










2 Answers
2






active

oldest

votes


















2












$begingroup$

The last expression can be written as
$$x_2(y_2−y_4)+a(y_4−y_3)+x_1(y_3−y_1)$$
Here all the terms are positive, hence the inequality follows.



Hope it is helpful:)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 12:25



















0












$begingroup$

If I am not mistaken, Mathematica answers no in such a way:



ForAll[{x1, x2, x3, x4, y1, y2, y3, y4, a}, 

 Implies[x1 < x3 && x3 < x4 && x4 < x2 && y1 < y3 && y3 < y4 &&  y4 < y2 && x1 + x2 == x3 + x4 && 

x1 + a == x3 && x2 - a == x4 && a > 0 && x1 > 0 && y1 > 0, x1*y1 + x2*y2 - (x3*y3 + x4*y4) > 0]];

Resolve[%, Reals]



False







share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The last expression can be written as
    $$x_2(y_2−y_4)+a(y_4−y_3)+x_1(y_3−y_1)$$
    Here all the terms are positive, hence the inequality follows.



    Hope it is helpful:)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
      $endgroup$
      – Thusitha Thilina Dayaratne
      Jan 17 at 12:25
















    2












    $begingroup$

    The last expression can be written as
    $$x_2(y_2−y_4)+a(y_4−y_3)+x_1(y_3−y_1)$$
    Here all the terms are positive, hence the inequality follows.



    Hope it is helpful:)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
      $endgroup$
      – Thusitha Thilina Dayaratne
      Jan 17 at 12:25














    2












    2








    2





    $begingroup$

    The last expression can be written as
    $$x_2(y_2−y_4)+a(y_4−y_3)+x_1(y_3−y_1)$$
    Here all the terms are positive, hence the inequality follows.



    Hope it is helpful:)






    share|cite|improve this answer









    $endgroup$



    The last expression can be written as
    $$x_2(y_2−y_4)+a(y_4−y_3)+x_1(y_3−y_1)$$
    Here all the terms are positive, hence the inequality follows.



    Hope it is helpful:)







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 17 at 12:06









    MartundMartund

    1,645213




    1,645213












    • $begingroup$
      I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
      $endgroup$
      – Thusitha Thilina Dayaratne
      Jan 17 at 12:25


















    • $begingroup$
      I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
      $endgroup$
      – Thusitha Thilina Dayaratne
      Jan 17 at 12:25
















    $begingroup$
    I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 12:25




    $begingroup$
    I think I made a mistake in the last statement. I corrected it now in the question. It should be x1(y1−y3). But since y1 -y3 < 0 we can't say the full statement is greater than 0.Sorry for the inconvenience
    $endgroup$
    – Thusitha Thilina Dayaratne
    Jan 17 at 12:25











    0












    $begingroup$

    If I am not mistaken, Mathematica answers no in such a way:



    ForAll[{x1, x2, x3, x4, y1, y2, y3, y4, a}, 
    
 Implies[x1 < x3 && x3 < x4 && x4 < x2 && y1 < y3 && y3 < y4 &&  y4 < y2 && x1 + x2 == x3 + x4 && 
    
x1 + a == x3 && x2 - a == x4 && a > 0 && x1 > 0 && y1 > 0, x1*y1 + x2*y2 - (x3*y3 + x4*y4) > 0]];
    
Resolve[%, Reals]



    False







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If I am not mistaken, Mathematica answers no in such a way:



      ForAll[{x1, x2, x3, x4, y1, y2, y3, y4, a}, 
      
 Implies[x1 < x3 && x3 < x4 && x4 < x2 && y1 < y3 && y3 < y4 &&  y4 < y2 && x1 + x2 == x3 + x4 && 
      
x1 + a == x3 && x2 - a == x4 && a > 0 && x1 > 0 && y1 > 0, x1*y1 + x2*y2 - (x3*y3 + x4*y4) > 0]];
      
Resolve[%, Reals]



      False







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If I am not mistaken, Mathematica answers no in such a way:



        ForAll[{x1, x2, x3, x4, y1, y2, y3, y4, a}, 
        
 Implies[x1 < x3 && x3 < x4 && x4 < x2 && y1 < y3 && y3 < y4 &&  y4 < y2 && x1 + x2 == x3 + x4 && 
        
x1 + a == x3 && x2 - a == x4 && a > 0 && x1 > 0 && y1 > 0, x1*y1 + x2*y2 - (x3*y3 + x4*y4) > 0]];
        
Resolve[%, Reals]



        False







        share|cite|improve this answer









        $endgroup$



        If I am not mistaken, Mathematica answers no in such a way:



        ForAll[{x1, x2, x3, x4, y1, y2, y3, y4, a}, 
        
 Implies[x1 < x3 && x3 < x4 && x4 < x2 && y1 < y3 && y3 < y4 &&  y4 < y2 && x1 + x2 == x3 + x4 && 
        
x1 + a == x3 && x2 - a == x4 && a > 0 && x1 > 0 && y1 > 0, x1*y1 + x2*y2 - (x3*y3 + x4*y4) > 0]];
        
Resolve[%, Reals]



        False








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 13:22









        user64494user64494

        2,971932




        2,971932






























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