repost-old-question-unanswer-Given-4x-2-1-4x-2-y-4y-2-1-4y-2-z-4z-2-1-4z-2-x-

Question Number 106794 by bobhans last updated on 07/Aug/20

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 @}

{\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

\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