Can the System of Inequalities in R be Solved with Real Numbers x, y, and z?

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SUMMARY

The system of inequalities involving real numbers x, y, and z can be solved using the equations (1+4x²)y=4z², (1+4y²)z=4x², and (1+4z²)x=4y². A key finding is that the solution (1/2, 1/2, 1/2) satisfies the equations, along with the trivial solution (0, 0, 0). The discussion highlights the complexity of solving these equations, particularly when substituting values and rearranging terms to isolate variables.

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Homework Statement


x y and z are real numbers
solve in lR
(1+4x^2)y=4z^2
(1+4y^2)z=4x^2
(1+4z^2)x=4y^22. The attempt at a solution
(1+4x^2)y(1+4y^2)z(1+4z^2)x=64x^2y^2z^2(x and y and z =/=0)
(\frac{1}{4}+x^2)(\frac{1}{4}+y^2)(\frac{1}{4}+z^2)=xyz
now i can tell from this that (1/2,1/2,1/2)(suppose that x=y=z solves this also (0,0,0,) is a solution)
but I'm really stuck here i thought this problem would be easy i need help
 
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I would try this: plugging (1) in (3), (1+(1+4x^2)y)x=4y^2 or 4y^2-(x+4x^3)y-x=0
This can be solved for y.
 

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