Given { ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z)),((((4z^2 )/(4z^2 +1)) = x)) :} . find x+y+z ?


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Question Number 105448 by bemath last updated on 29/Jul/20
$Given { ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z)),((((4z^2 )/(4z^2 +1)) = x)) :} . find x+y+z ?$
$G i v e n {\begin{cases} \frac{4 x^{2}}{4 x^{2} + 1} = y \\ \frac{4 y^{2}}{4 y^{2} + 1} = z \\ \frac{4 z^{2}}{4 z^{2} + 1} = x \end{cases} . f i n d$ $x + y + z ?$
Commented by john santu last updated on 29/Jul/20
$(1/x)+(1/y)+(1/z)= ((4x^2 +1)/(4x^2 ))+((4y^2 +1)/(4y^2 ))+((4z^2 +1)/(4z^2 )) (1/x)+(1/y)+(1/z)= 3+(1/(4x^2 ))+(1/(4y^2 ))+(1/(4z^2 )) (1/x)+(1/y)+(1/z)=3+((1/(2x)))^2 +((1/(2y)))^2 +((1/(2z)))^2 (1/x)+(1/y)+(1/z)=3+(1/4){((1/x))^2 +((1/y))^2 +((1/z))^2 } { ((set (1/x)=a, (1/y)=b, (1/z)=c {: (),() })) :} ⇔ a+b+c = 3 +(1/4){(a+b+c)^2 −2(ab+ac+bc)}$
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{4 x^{2} + 1}{4 x^{2}} + \frac{4 y^{2} + 1}{4 y^{2}} + \frac{4 z^{2} + 1}{4 z^{2}}$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + \frac{1}{4 x^{2}} + \frac{1}{4 y^{2}} + \frac{1}{4 z^{2}}$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + {(\frac{1}{2 x})}^{2} + {(\frac{1}{2 y})}^{2} + {(\frac{1}{2 z})}^{2}$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 + \frac{1}{4} {{(\frac{1}{x})}^{2} + {(\frac{1}{y})}^{2} + {(\frac{1}{z})}^{2}}$ ${\begin{cases} s e t \frac{1}{x} = a, \frac{1}{y} = b, \frac{1}{z} = c \begin{array}{c} \end{array}} \end{cases}$ $\Leftrightarrow a + b + c = 3 + \frac{1}{4} {{(a + b + c)}^{2} - 2 (a b + a c + b c)}$
Commented by behi83417@gmail.com last updated on 29/Jul/20
$x=y=z=0 and:4x^2 −4x+1=0⇒x=y=z=(1/2) if:x≠y≠z≠0⇒x=(1/(2a)),y=(1/(2b)),z=(1/(2c))⇒ { ((((4.(1/(4a^2 )))/(4.(1/(4a^2 ))+1))=(1/(2b))⇒ { (((1/(1+a^2 ))=(1/(2b)))),(((1/(1+b^2 ))=(1/(2c))⇒)) :})),() :} ⇒ { ((a^2 +1=2b)),((b^2 +1=2c)),((c^2 +1=2a)) :}⇒(a−1)^2 +(b−1)^2 +(c−1)^2 =0 ⇒a=b=c=1⇒x=y=z=(1/2)$
$x = y = z = 0 and : {4 x}^{2} - 4 x + 1 = 0 \Rightarrow x = y = z = \frac{1}{2}$ $if : x \neq y \neq z \neq 0 \Rightarrow x = \frac{1}{2 a}, y = \frac{1}{2 b}, z = \frac{1}{2 c} \Rightarrow$ ${\begin{cases} \frac{4 . \frac{1}{{4 a}^{2}}}{4 . \frac{1}{{4 a}^{2}} + 1} = \frac{1}{2 b} \Rightarrow {\begin{cases} \frac{1}{1 + a^{2}} = \frac{1}{2 b} \\ \frac{1}{1 + b^{2}} = \frac{1}{2 c} \Rightarrow \end{cases} \end{cases}$ $\Rightarrow {\begin{cases} a^{2} + 1 = 2 b \\ b^{2} + 1 = 2 c \\ c^{2} + 1 = 2 a \end{cases} \Rightarrow {(a - 1)}^{2} + {(b - 1)}^{2} + {(c - 1)}^{2} = 0$ $\Rightarrow a = b = c = 1 \Rightarrow x = y = z = \frac{1}{2}$
	Answered by ajfour last updated on 29/Jul/20

	$l e t \frac{1}{x} = p, \frac{1}{y} = q, \frac{1}{z} = r$ $\Rightarrow \frac{4}{4 - p^{2}} = \frac{1}{q}, \frac{4}{4 {(\frac{q}{p})}^{2} + q^{2}} = \frac{1}{r},$ $\frac{4}{4 + r^{2}} = \frac{1}{p}$ $\Rightarrow 4 - p^{2} = 4 q . . . (i)$ $4 {(\frac{q}{p})}^{2} + q^{2} = 4 r \Rightarrow q^{2} (\frac{1}{p^{2}} + \frac{1}{4}) = r . . (i i)$ $4 + r^{2} = 4 p . . . (i i i)$ $\Rightarrow {(1 - \frac{p^{2}}{4})}^{2} {(\frac{1}{p^{2}} + \frac{1}{4})}^{2} = 4 (p - 1)$ $\Rightarrow {(16 - p^{4})}^{2} = 64 \times 16 p^{4} (p - 1)$ $. . . . . . d e g r e e 8$
	Answered by bramlex last updated on 29/Jul/20
	$(1/y) = ((4x^2 +1)/(4x^2 )) →(1/y) = 1+(1/(4x^2 )) →(1/z) = 1+(1/4)(1+(1/(4x^2 )))^2 →1+(1/(4z^2 )) = (1/x) →1+(1/4){1+(1/4)(1+(1/(4x^2 )))^2 }= (1/x) set (1/x) = m 1+(1/4){1+(1/4)(1+(1/4)m^2 )^2 } = m (5/4)+(1/4)(1+(1/2)m^2 +(1/(16))m^4 )=m 20+4+2m^2 +(1/4)m^4 =16m m^4 +32m^2 −256m+16×24=0$
	$\frac{1}{y} = \frac{4 x^{2} + 1}{4 x^{2}} \to \frac{1}{y} = 1 + \frac{1}{4 x^{2}}$ $\to \frac{1}{z} = 1 + \frac{1}{4} {(1 + \frac{1}{4 x^{2}})}^{2}$ $\to 1 + \frac{1}{4 z^{2}} = \frac{1}{x}$ $\to 1 + \frac{1}{4} {1 + \frac{1}{4} {(1 + \frac{1}{4 x^{2}})}^{2}} = \frac{1}{x}$ $s e t \frac{1}{x} = m$ $1 + \frac{1}{4} {1 + \frac{1}{4} {(1 + \frac{1}{4} m^{2})}^{2}} = m$ $\frac{5}{4} + \frac{1}{4} (1 + \frac{1}{2} m^{2} + \frac{1}{16} m^{4}) = m$ $20 + 4 + 2 m^{2} + \frac{1}{4} m^{4} = 16 m$ $m^{4} + 32 m^{2} - 256 m + 16 \times 24 = 0$
	Answered by 1549442205PVT last updated on 07/Aug/20
	$WLOG suppose that x≥y≥z≥0 (1).Then ((4z^2 )/(4z^2 +1))≥((4x^2 )/(4x^2 +1))≥((4y^2 )/(4y^2 +1))≥0 ⇒(z^2 /(4z^2 +1))≥(x^2 /(4x^2 +1))⇒z^2 (4x^2 +1)≥x^2 (4z^2 +1) ⇒z^2 ≥x^2 ⇒z≥x≥0(2).From (1) and(2) we get x=y=z.Hence ((4x^2 )/(4x^2 +1))=x⇔4x^3 −4x^2 +x=0 ⇔x(4x^2 −4x+1)=0.⇔x(2x−1)^2 =0 ⇔x∈{0;(1/2)} Thus,the roots of the given system are: {(0;0);((1/2);(1/2))}$
	$WLOG suppose that x ⩾ y ⩾ z ⩾ 0 (1) . Then$ $\frac{{4 z}^{2}}{{4 z}^{2} + 1} ⩾ \frac{{4 x}^{2}}{{4 x}^{2} + 1} ⩾ \frac{{4 y}^{2}}{{4 y}^{2} + 1} ⩾ 0$ $\Rightarrow \frac{z^{2}}{{4 z}^{2} + 1} ⩾ \frac{x^{2}}{{4 x}^{2} + 1} \Rightarrow z^{2} ({4 x}^{2} + 1) ⩾ x^{2} ({4 z}^{2} + 1)$ $\Rightarrow z^{2} ⩾ x^{2} \Rightarrow z ⩾ x ⩾ 0 (2) . From (1) and (2)$ $we get x = y = z . Hence$ $\frac{{4 x}^{2}}{{4 x}^{2} + 1} = x \Leftrightarrow {4 x}^{3} - {4 x}^{2} + x = 0$ $\Leftrightarrow x ({4 x}^{2} - 4 x + 1) = 0 . \Leftrightarrow x {(2 x - 1)}^{2} = 0$ $\Leftrightarrow x \in {0; \frac{1}{2}}$ $Thus, the roots of the given system are :$ ${(0; 0); (\frac{1}{2}; \frac{1}{2})}$
	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

	$Thank Sir . I understanded and shall$ $correct it$
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