repost old question unanswer Given → { ((((4x^2 )/(1+4x^2 )) = y)),((((4y^2 )/(1+4y^2 )) = z)),((((4z^2 )/(1+4z^2 )) = x)) :}


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Question Number 106794 by bobhans last updated on 07/Aug/20
$repost old question unanswer Given → { ((((4x^2 )/(1+4x^2 )) = y)),((((4y^2 )/(1+4y^2 )) = z)),((((4z^2 )/(1+4z^2 )) = x)) :}$
$repost old question unanswer$ $G iven \to {\begin{cases} \frac{{4 x}^{2}}{1 + {4 x}^{2}} = y \\ \frac{{4 y}^{2}}{1 + {4 y}^{2}} = z \\ \frac{{4 z}^{2}}{1 + {4 z}^{2}} = x \end{cases}$
	Answered by thearith last updated on 07/Aug/20


	Answered by john santu last updated on 07/Aug/20
	$^(@JS@) { ((1+(1/(4x^2 )) = (1/y))),((1+(1/(4y^2 )) = (1/z))),((1+(1/(4z^2 )) = (1/x))) :} ⇒(1/(4x^2 ))+(1/(4y^2 ))+(1/(4z^2 ))+3 = (1/x)+(1/y)+(1/z) ⇒((1/(4x^2 ))−(1/x)+1)+((1/(4y^2 ))−(1/y)+1)+((1/(4z^2 ))−(1/z)+1)=0 (((4x^2 −4x+1)/(4x^2 )))+(((4y^2 −4y+1)/(4y^2 )))+(((4z^2 −4z+1)/(4z^2 )))=0 (((2x−1)/(2x)))^2 +(((2y−1)/(2y)))^2 +(((2z−1)/(2z)))^2 =0 { ((x = (1/2))),((y=(1/2))),((z=(1/2))) :}.$
	$^{@ JS @}$ ${\begin{cases} 1 + \frac{1}{{4 x}^{2}} = \frac{1}{y} \\ 1 + \frac{1}{{4 y}^{2}} = \frac{1}{z} \\ 1 + \frac{1}{{4 z}^{2}} = \frac{1}{x} \end{cases}$ $\Rightarrow \frac{1}{{4 x}^{2}} + \frac{1}{{4 y}^{2}} + \frac{1}{{4 z}^{2}} + 3 = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ $\Rightarrow (\frac{1}{{4 x}^{2}} - \frac{1}{x} + 1) + (\frac{1}{{4 y}^{2}} - \frac{1}{y} + 1) + (\frac{1}{{4 z}^{2}} - \frac{1}{z} + 1) = 0$ $(\frac{{4 x}^{2} - 4 x + 1}{{4 x}^{2}}) + (\frac{{4 y}^{2} - 4 y + 1}{{4 y}^{2}}) + (\frac{{4 z}^{2} - 4 z + 1}{{4 z}^{2}}) = 0$ ${(\frac{2 x - 1}{2 x})}^{2} + {(\frac{2 y - 1}{2 y})}^{2} + {(\frac{2 z - 1}{2 z})}^{2} = 0$ ${\begin{cases} x = \frac{1}{2} \\ y = \frac{1}{2} \\ z = \frac{1}{2} \end{cases} .$
	Commented by bemath last updated on 07/Aug/20

	$thank you$
	Commented by bobhans last updated on 07/Aug/20

	$and the other solution {\begin{cases} x = 0 \\ y = 0 \\ z = 0 \end{cases}$
	Answered by 1549442205PVT last updated on 07/Aug/20
	$WLOG suppose that x≥y≥z≥0.Then from the hypothesis we have ((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ≥((4y^2 )/(1+4y^2 )) ⇒((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ⇔4z^2 (1+4x^2 )≥4x^2 (1+4z^2 )⇔z^2 ≥x^2 ⇒x=y=z.Replace into (1) we get 4x^2 =x(1+4x^2 )⇔4x^3 −4x^2 +x=0 ⇔x(4x^2 −4x+1)=0⇔x(2x−1)^2 =0 ⇔x∈{0;(1/2)} Hence,the given system has two solutions as {(x,y,z)∈{(0,0,0);((1/2);(1/2),(1/2))}$
	$WLOG suppose that x ⩾ y ⩾ z ⩾ 0 . Then$ $from the hypothesis we have$ $\frac{{4 z}^{2}}{1 + {4 z}^{2}} ⩾ \frac{{4 x}^{2}}{1 + {4 x}^{2}} ⩾ \frac{{4 y}^{2}}{1 + {4 y}^{2}} \Rightarrow \frac{{4 z}^{2}}{1 + {4 z}^{2}} ⩾ \frac{{4 x}^{2}}{1 + {4 x}^{2}}$ $\Leftrightarrow {4 z}^{2} (1 + {4 x}^{2}) ⩾ {4 x}^{2} (1 + {4 z}^{2}) \Leftrightarrow z^{2} ⩾ x^{2}$ $\Rightarrow x = y = z . Replace into (1) we get$ ${4 x}^{2} = x (1 + {4 x}^{2}) \Leftrightarrow {4 x}^{3} - {4 x}^{2} + x = 0$ $\Leftrightarrow x ({4 x}^{2} - 4 x + 1) = 0 \Leftrightarrow x {(2 x - 1)}^{2} = 0$ $\Leftrightarrow x \in {0; \frac{1}{2}} Hence, the given system$ $has two solutions as$ ${(x, y, z) \in {(0, 0, 0); (\frac{1}{2}; \frac{1}{2}, \frac{1}{2})}$
	Commented by bemath last updated on 07/Aug/20

	$sir what is WLOG$
	Commented by 1549442205PVT last updated on 07/Aug/20

	$it is the brief writting of phrase$ $‘ ‘ without loss the {generality}^{″} is used$ $in the math . proof$
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