4x-2-4x-2-1-y-4y-2-4y-2-1-z-4z-2-4z-2-1-x-where-x-y-z-0-

Question Number 120110 by benjo_mathlover last updated on 29/Oct/20

{\begin{cases} \frac{4 x^{2}}{4 x^{2} + 1} = y \\ \frac{4 y^{2}}{4 y^{2} + 1} = z \\ \frac{4 z^{2}}{4 z^{2} + 1} = x \end{cases}

w h e r e x, y, z \neq 0

Answered by bemath last updated on 29/Oct/20

\Rightarrow {\begin{cases} \frac{1}{y} = 1 + \frac{1}{4 x^{2}} \\ \frac{1}{z} = 1 + \frac{1}{4 y^{2}} \\ \frac{1}{x} = 1 + \frac{1}{4 z^{2}} \end{cases}

(1 + \frac{1}{4 x^{2}} - \frac{1}{x}) + (1 + \frac{1}{4 y^{2}} - \frac{1}{y}) + (1 + \frac{1}{4 z} - \frac{1}{z}) = 0

{(1 - \frac{1}{2 x})}^{2} + {(1 - \frac{1}{2 y})}^{2} + {(1 - \frac{1}{2 z})}^{2} = 0

\to \frac{1}{2 x} = \frac{1}{2 y} = \frac{1}{2 z} = 1, (x, y, z) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})

Answered by mindispower last updated on 29/Oct/20

y = f (4 x^{2}), f (t) = \frac{t}{t + 1}, f^{'} (t) ⩾ 0, f i n c r e a s e

x, y, z ⩾ 0

z = f 0 f (4 x^{2})

x = f o f o f (4 x^{2}),

f o f i n c r e a s e f o n c t i o n

4 x^{2} > 4 y^{2} \Rightarrow f (4 x^{2}) > f (4 y^{2}) \Rightarrow y > z \Rightarrow 4 y^{2} > 4 z^{2}

\Rightarrow f (4 y^{2}) > f (4 z^{2}) \Leftrightarrow z > x

z < y \Rightarrow x > y > z > x a b s u r d e x < y a l s o \Rightarrow x > y

s o x = y \Rightarrow x = \frac{4 x^{2}}{4 x^{2} + 1} \Leftrightarrow x (4 x^{2} - 4 x + 1) = 0

x {(2 x - 1)}^{2} = 0 \Leftrightarrow x = 0, x = \frac{1}{2}

S = (x, y, z) \in {(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})}