MHB Find the condition for equality to hold

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The discussion revolves around proving the condition for equality in the equation x√(1-y²) + y√(1-x²) = 1, specifically that x² + y² = 1. The Cauchy-Schwarz inequality is utilized to show that the left-hand side is less than or equal to 1. It is noted that equality holds when specific ratios involving x and y are satisfied. The user reflects on their approach and acknowledges missing a crucial part of the definition related to equality conditions. The conversation emphasizes the relationship between the original equation and the condition x² + y² = 1.
anemone
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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.
 
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anemone said:
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:
Gosh, I missed that part of definition!:o

Thanks, Alexmahone.
 
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