Welcome to our community

Be a part of something great, join today!

Find the condition for equality to hold

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,596
Hi,
I've encountered a problem in deciding the condition in order for the equality to hold.
Here is the problem:

If $x\sqrt {1-y^2} + y \sqrt {1-x^2}=1$, prove that $x^2+y^2=1$

By using the Cauchy-Schwarz inequality, it's fairly easy to prove that $x\sqrt {1-y^2} + y \sqrt {1-x^2}\leq1$

Next, what I tried to do is to work backwards and let $x^2+y^2=1$, then I see that $x=\sqrt {1-y^2}$. After making that substitution into the LHS of the inequality $ x\sqrt {1-y^2} + y \sqrt {1-x^2} $ and I eventually get 1 as the final answer.

What do you think, Sir? I feel bad for doing this.

Do you have any idea to deduce the condition from $x\sqrt {1-y^2} + y \sqrt {1-x^2}\leq1$?

Thanks.
 

Alexmahone

Active member
Jan 26, 2012
268
Hi,
I've encountered a problem in deciding the condition in order for the equality to hold.
Here is the problem:

If $x\sqrt {1-y^2} + y \sqrt {1-x^2}=1$, prove that $x^2+y^2=1$

By using the Cauchy-Schwarz inequality, it's fairly easy to prove that $x\sqrt {1-y^2} + y \sqrt {1-x^2}\leq1$
In the Cauchy Schwarz inequality, equality holds only if

$\displaystyle \frac{x}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{y}$

$\displaystyle xy=\sqrt{(1-x^2)(1-y^2)}$

$\displaystyle x^2y^2=(1-x^2)(1-y^2)=1-x^2-y^2+x^2y^2$

$\displaystyle x^2+y^2=1$
 
Last edited:
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,596
Gosh, I missed that part of definition!:eek:

Thanks, Alexmahone.