Given-4x-2-4x-2-1-y-4y-2-4y-2-1-z-4z-2-4z-2-1-x-find-x-y-z-

Question Number 105448 by bemath last updated on 29/Jul/20

G i v e n {\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} . f i n d

x + y + z ?

Commented by john santu last updated on 29/Jul/20

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

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + \frac{1}{4 x^{2}} + \frac{1}{4 y^{2}} + \frac{1}{4 z^{2}}

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + {(\frac{1}{2 x})}^{2} + {(\frac{1}{2 y})}^{2} + {(\frac{1}{2 z})}^{2}

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + \frac{1}{4} {{(\frac{1}{x})}^{2} + {(\frac{1}{y})}^{2} + {(\frac{1}{z})}^{2}}

{\begin{cases} s e t \frac{1}{x} = a, \frac{1}{y} = b, \frac{1}{z} = c \begin{array}{c}  \end{array}} \end{cases}

\Leftrightarrow a + b + c = 3 + \frac{1}{4} {{(a + b + c)}^{2} - 2 (a b + a c + b c)}

Commented by behi83417@gmail.com last updated on 29/Jul/20

x = y = z = 0 and : {4 x}^{2} - 4 x + 1 = 0 \Rightarrow x = y = z = \frac{1}{2}

if : x \neq y \neq z \neq 0 \Rightarrow x = \frac{1}{2 a}, y = \frac{1}{2 b}, z = \frac{1}{2 c} \Rightarrow

{\begin{cases} \frac{4 . \frac{1}{{4 a}^{2}}}{4 . \frac{1}{{4 a}^{2}} + 1} = \frac{1}{2 b} \Rightarrow {\begin{cases} \frac{1}{1 + a^{2}} = \frac{1}{2 b} \\ \frac{1}{1 + b^{2}} = \frac{1}{2 c} \Rightarrow \end{cases} \end{cases}

\Rightarrow {\begin{cases} a^{2} + 1 = 2 b \\ b^{2} + 1 = 2 c \\ c^{2} + 1 = 2 a \end{cases} \Rightarrow {(a - 1)}^{2} + {(b - 1)}^{2} + {(c - 1)}^{2} = 0

\Rightarrow a = b = c = 1 \Rightarrow x = y = z = \frac{1}{2}

Answered by ajfour last updated on 29/Jul/20

l e t \frac{1}{x} = p, \frac{1}{y} = q, \frac{1}{z} = r

\Rightarrow \frac{4}{4 - p^{2}} = \frac{1}{q}, \frac{4}{4 {(\frac{q}{p})}^{2} + q^{2}} = \frac{1}{r},

\frac{4}{4 + r^{2}} = \frac{1}{p}

\Rightarrow 4 - p^{2} = 4 q \dots (i)

4 {(\frac{q}{p})}^{2} + q^{2} = 4 r \Rightarrow q^{2} (\frac{1}{p^{2}} + \frac{1}{4}) = r . . (i i)

4 + r^{2} = 4 p \dots (i i i)

\Rightarrow {(1 - \frac{p^{2}}{4})}^{2} {(\frac{1}{p^{2}} + \frac{1}{4})}^{2} = 4 (p - 1)

\Rightarrow {(16 - p^{4})}^{2} = 64 \times 16 p^{4} (p - 1)

\dots \dots d e g r e e 8

Answered by bramlex last updated on 29/Jul/20

\frac{1}{y} = \frac{4 x^{2} + 1}{4 x^{2}} \to \frac{1}{y} = 1 + \frac{1}{4 x^{2}}

\to \frac{1}{z} = 1 + \frac{1}{4} {(1 + \frac{1}{4 x^{2}})}^{2}

\to 1 + \frac{1}{4 z^{2}} = \frac{1}{x}

\to 1 + \frac{1}{4} {1 + \frac{1}{4} {(1 + \frac{1}{4 x^{2}})}^{2}} = \frac{1}{x}

s e t \frac{1}{x} = m

1 + \frac{1}{4} {1 + \frac{1}{4} {(1 + \frac{1}{4} m^{2})}^{2}} = m

\frac{5}{4} + \frac{1}{4} (1 + \frac{1}{2} m^{2} + \frac{1}{16} m^{4}) = m

20 + 4 + 2 m^{2} + \frac{1}{4} m^{4} = 16 m

m^{4} + 32 m^{2} - 256 m + 16 \times 24 = 0

Answered by 1549442205PVT last updated on 07/Aug/20

WLOG suppose that x ⩾ y ⩾ z ⩾ 0 (1) . Then

\frac{{4 z}^{2}}{{4 z}^{2} + 1} ⩾ \frac{{4 x}^{2}}{{4 x}^{2} + 1} ⩾ \frac{{4 y}^{2}}{{4 y}^{2} + 1} ⩾ 0

\Rightarrow \frac{z^{2}}{{4 z}^{2} + 1} ⩾ \frac{x^{2}}{{4 x}^{2} + 1} \Rightarrow z^{2} ({4 x}^{2} + 1) ⩾ x^{2} ({4 z}^{2} + 1)

\Rightarrow z^{2} ⩾ x^{2} \Rightarrow z ⩾ x ⩾ 0 (2) . From (1) and (2)

we get x = y = z . Hence

\frac{{4 x}^{2}}{{4 x}^{2} + 1} = x \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}}

Thus, the roots of the given system are :

{(0; 0); (\frac{1}{2}; \frac{1}{2})}

Commented by behi83417@gmail.com last updated on 31/Jul/20

You can't use 'macro parameter character #' in math mode

Commented by 1549442205PVT last updated on 01/Aug/20

Thank Sir . I understanded and shall

correct it