MHB Finding the Ratio of x,y,z in $R^+$

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The discussion focuses on finding the ratio of positive real numbers x, y, and z defined by a set of equations involving square roots and constants. The equations express x, y, and z in terms of each other, incorporating subtractions of fractions from their squares. Participants are tasked with solving these equations to determine the proportional relationship between x, y, and z. The hint emphasizes that all variables are positive real numbers. The goal is to derive the ratio x:y:z from the given relationships.
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$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}$
$find \,\,\,\, x:y:z$
 
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Albert said:
$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}---(1)$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}---(2)$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}--(3)$
$find \,\,\,\, x:y:z$
hint:
eliminate square root sign for (1) (2) (3) and rearrange,to make the left sides of them all equal
 
Albert said:
$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}---(1)$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}---(2)$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}---(3)$
$find \,\,\,\, x:y:z$

eliminate square root sign and rearrange for (1)we get:
$(x-\sqrt{{y^2}-\dfrac{1}{25}})^2=(\sqrt{{z^2}-\dfrac{1}{25}})^2$
$\rightarrow x^2+y^2-z^2=2x\sqrt {y^2-\dfrac {1}{25}}$ (square both sides again)
$\rightarrow x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=\dfrac {-4x^2}{25}---(A)$
the same procedures for (2)and(3) we get :
$\rightarrow x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=\dfrac {-4y^2}{36}---(B)$
$\rightarrow x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=\dfrac {-4z^2}{49}---(C)$
from (A)(B)(C)we have:
$\dfrac {x^2}{5^2}=\dfrac {y^2}{6^2}=\dfrac {z^2}{7^2}$
$\therefore x:y:z=5:6:7(for\,\, x,y,z>0)$
 
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