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Question Number 109369 by I want to learn more last updated on 23/Aug/20

	Answered by floor(10²Eta[1]) last updated on 23/Aug/20

	${(x + y + z)}^{2} = 4 = x^{2} + y^{2} + z^{2} + 2 (xy + xz + yz)$ $\Rightarrow xy + xz + yz = \frac{5}{4}$ $xy + xz + yz = \frac{5}{4} = \frac{1}{4} (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ $\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 5 = \frac{x + y}{xy} + \frac{1}{z} = \frac{2 - z}{\frac{1}{4 z}} + \frac{1}{z}$ $\Rightarrow 8 z - {4 z}^{2} + \frac{1}{z} = 5 \Rightarrow {4 z}^{3} - {8 z}^{2} + 5 z - 1 = 0$ $z = 1 is a root ∴$ ${4 z}^{3} - {8 z}^{2} + 5 z - 1 = (z - 1) ({4 z}^{2} + bz + 1)$ $= {4 z}^{3} + (b - 4) z^{2} + (1 - b) z - 1 \Rightarrow b = - 4$ ${4 z}^{2} - 4 z + 1 = 0 \Rightarrow z = \frac{1}{2}$ $(1) : z = 1$ $x + y = 1$ $x^{2} + y^{2} = \frac{1}{2}$ $xy = \frac{1}{4}$ $\Rightarrow x = \frac{1}{2} ∴ y = \frac{1}{2}$ $(2) : z = \frac{1}{2}$ $x + y = \frac{3}{2}$ $x^{2} + y^{2} = \frac{5}{4}$ $xy = \frac{1}{2}$ $\Rightarrow x = 1 \lor x = \frac{1}{2} ∴ y = \frac{1}{2} \lor y = 1$ $so all solutions are :$ $(x, y, z) = (\frac{1}{2}, \frac{1}{2}, 1), (1, \frac{1}{2}, \frac{1}{2}), (\frac{1}{2}, 1, \frac{1}{2})$
	Commented by I want to learn more last updated on 23/Aug/20

	$Thanks sir$
	Answered by 1549442205PVT last updated on 23/Aug/20
	${ ((x^2 +y^2 +z^2 =3/2)),((x+y+z=2)) :} ⇒xy+yz+zx=(((x+y+z)^2 −(x^2 +y^2 +z^2 ))/2) =((2^2 −3/2)/2)=(5/4) { ((x+y+z=2)),((xy+yz+zx=5/4)),((xyz=1/4)) :} From Vieta′s theorem we infer that x,y,z are the roots of the equation of degree 3: t^3 −2t^2 +(5/4)t−(1/4)=0 ⇔4t^3 −8t^2 +5t−1=0⇔(t−1)(4t^2 −4t+1)=0 ⇔(t−1)(2t−1)=0⇒t_(1,2,3) ∈{1,(1/2),(1/2)} ⇒(x,y,z)∈{(1,(1/2),(1/2)),((1/2),1,(1/2)),((1/2),(1/2),1)}$
	${\begin{cases} x^{2} + y^{2} + z^{2} = 3 / 2 \\ x + y + z = 2 \end{cases}$ $\Rightarrow xy + yz + zx = \frac{{(x + y + z)}^{2} - (x^{2} + y^{2} + z^{2})}{2}$ $= \frac{2^{2} - 3 / 2}{2} = \frac{5}{4}$ ${\begin{cases} x + y + z = 2 \\ xy + yz + zx = 5 / 4 \\ xyz = 1 / 4 \end{cases}$ $From {Vieta}^{'} s theorem we infer that$ $x, y, z are the roots of the equation$ $of degree 3 : t^{3} - {2 t}^{2} + \frac{5}{4} t - \frac{1}{4} = 0$ $\Leftrightarrow {4 t}^{3} - {8 t}^{2} + 5 t - 1 = 0 \Leftrightarrow (t - 1) ({4 t}^{2} - 4 t + 1) = 0$ $\Leftrightarrow (t - 1) (2 t - 1) = 0 \Rightarrow t_{1, 2, 3} \in {1, \frac{1}{2}, \frac{1}{2}}$ $\Rightarrow (x, y, z) \in {(1, \frac{1}{2}, \frac{1}{2}), (\frac{1}{2}, 1, \frac{1}{2}), (\frac{1}{2}, \frac{1}{2}, 1)}$
	Commented by I want to learn more last updated on 23/Aug/20

	$Thanks sir$
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