Equation & Inequation

Algebra Level 4

{ x + y 1 2 x y + 1 x 2 + y 2 = 10 \begin{cases} x+y \leq 1 \\ \dfrac2{xy} + \dfrac1{x^2+y^2} = 10 \end{cases}

Find x x and y y such that they satisfy the constraints above. Submit your answer as x + y |x+y| .

If you think there are no x x and y y satisfying these equations, enter 123456 as your answer.


The answer is 1.

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2 solutions

Shaun Leong
Feb 1, 2016

I agree that x and y should be positive reals. Note that by AM-GM Inequality x y ( x + y 2 ) 2 1 4 xy \leq (\frac {x+y}{2})^2 \leq \frac {1}{4} x 2 + y 2 = ( x + y ) 2 2 x y 1 2 2 1 4 = 1 2 x^2+y^2 = (x+y)^2-2xy \leq 1^2-2*\frac {1}{4} =\frac {1}{2} with equality when x = y = 1 2 x=y=\frac {1}{2}

10 = 2 x y + 1 x 2 + y 2 2 1 4 + 1 1 2 = 10 10=\frac {2}{xy}+\frac {1}{x^2+y^2} \geq \frac {2}{\frac {1}{4}}+\frac {1}{\frac {1}{2}}=10 x = y = 1 2 \Rightarrow x=y=\frac {1}{2 }

Hence x + y = 1 |x+y|=\boxed{1}

P C
Jan 27, 2016

I think you should add "x, y are positive reals"

What if x = y = -1/2?

Siva Bathula - 4 years ago

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