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

[Albert1]

$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}$
$find \,\,\,\, x:y:z$

 

[Albert1]

$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}---(1)$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}---(2)$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}--(3)$
$find \,\,\,\, x:y:z$

hint:

Spoiler

eliminate square root sign for (1) (2) (3) and rearrange,to make the left sides of them all equal

 

[Albert1]

$x,y,z\in R^+$
$x=\sqrt{{y^2}-\dfrac{1}{25}}+\sqrt{{z^2}-\dfrac{1}{25}}---(1)$
$y=\sqrt{{z^2}-\dfrac{1}{36}}+\sqrt{{x^2}-\dfrac{1}{36}}---(2)$
$z=\sqrt{{x^2}-\dfrac{1}{49}}+\sqrt{{y^2}-\dfrac{1}{49}}---(3)$
$find \,\,\,\, x:y:z$

Spoiler

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:
$\dfrac {x^2}{5^2}=\dfrac {y^2}{6^2}=\dfrac {z^2}{7^2}$
$\therefore x:y:z=5:6:7(for\,\, x,y,z>0)$