SOLUTION: If a^2+b^2=1=c^2+d^2 where a,b,c,d are real numbers prove that ab+cd ≦ 1. (Use the result x^2+y^2=2xy) I have tried to prove this but I don't think my methods are right.

Question 805781: If a^2+b^2=1=c^2+d^2 where a,b,c,d are real numbers prove that ab+cd ≦ 1.
(Use the result x^2+y^2=2xy)
I have tried to prove this but I don't think my methods are right.
Any help would be greatly appreciated.
Thank you.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Something has been lost in transcription,
because was not true for nine out of ten sets of x, and y pairs I tried.
It is true, predictably, for ,
but while

I would prove that inequality like this:
-->
-->
Adding the two highlighted squares we get
-->
and since squares are non-negative, .
So,
--> --> -->
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SOLUTION: If a^2+b^2=1=c^2+d^2 where a,b,c,d are real numbers prove that ab+cd &#8806; 1. (Use the result x^2+y^2=2xy) I have tried to prove this but I don't think my methods are right.

SOLUTION: If a^2+b^2=1=c^2+d^2 where a,b,c,d are real numbers prove that ab+cd ≦ 1. (Use the result x^2+y^2=2xy) I have tried to prove this but I don't think my methods are right.