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Question Number 106794 by bobhans last updated on 07/Aug/20
repost old question unanswer Given → { ((((4x^2 )/(1+4x^2 )) = y)),((((4y^2 )/(1+4y^2 )) = z)),((((4z^2 )/(1+4z^2 )) = x)) :}
$$\mathrm{repost}\:\mathrm{old}\:\mathrm{question}\:\mathrm{unanswer} \\ $$$$\mathcal{G}\mathrm{iven}\:\rightarrow\begin{cases}{\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:=\:\mathrm{y}}\\{\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4y}^{\mathrm{2}} }\:=\:\mathrm{z}}\\{\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\:=\:\mathrm{x}}\end{cases} \\ $$
Answered by thearith last updated on 07/Aug/20
$$ \\ $$
Answered by john santu last updated on 07/Aug/20
^(@JS@) { ((1+(1/(4x^2 )) = (1/y))),((1+(1/(4y^2 )) = (1/z))),((1+(1/(4z^2 )) = (1/x))) :} ⇒(1/(4x^2 ))+(1/(4y^2 ))+(1/(4z^2 ))+3 = (1/x)+(1/y)+(1/z) ⇒((1/(4x^2 ))−(1/x)+1)+((1/(4y^2 ))−(1/y)+1)+((1/(4z^2 ))−(1/z)+1)=0 (((4x^2 −4x+1)/(4x^2 )))+(((4y^2 −4y+1)/(4y^2 )))+(((4z^2 −4z+1)/(4z^2 )))=0 (((2x−1)/(2x)))^2 +(((2y−1)/(2y)))^2 +(((2z−1)/(2z)))^2 =0 { ((x = (1/2))),((y=(1/2))),((z=(1/2))) :}.
$$\:\:\:\:^{@\mathrm{JS}@} \\ $$$$\begin{cases}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{y}}}\\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{z}}}\\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{x}}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }+\mathrm{3}\:=\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{y}}+\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{z}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{4y}^{\mathrm{2}} −\mathrm{4y}+\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{4z}^{\mathrm{2}} −\mathrm{4z}+\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2y}−\mathrm{1}}{\mathrm{2y}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2z}−\mathrm{1}}{\mathrm{2z}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\begin{cases}{\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases}. \\ $$
Commented by bemath last updated on 07/Aug/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by bobhans last updated on 07/Aug/20
and the other solution { ((x=0)),((y=0)),((z=0)) :}
$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{solution}\:\begin{cases}{\mathrm{x}=\mathrm{0}}\\{\mathrm{y}=\mathrm{0}}\\{\mathrm{z}=\mathrm{0}}\end{cases} \\ $$
Answered by 1549442205PVT last updated on 07/Aug/20
WLOG suppose that x≥y≥z≥0.Then from the hypothesis we have ((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ≥((4y^2 )/(1+4y^2 )) ⇒((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ⇔4z^2 (1+4x^2 )≥4x^2 (1+4z^2 )⇔z^2 ≥x^2 ⇒x=y=z.Replace into (1) we get 4x^2 =x(1+4x^2 )⇔4x^3 −4x^2 +x=0 ⇔x(4x^2 −4x+1)=0⇔x(2x−1)^2 =0 ⇔x∈{0;(1/2)} Hence,the given system has two solutions as {(x,y,z)∈{(0,0,0);((1/2);(1/2),(1/2))}
$$\mathrm{WLOG}\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{x}\geqslant\mathrm{y}\geqslant\mathrm{z}\geqslant\mathrm{0}.\mathrm{Then} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\geqslant\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:\geqslant\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4y}^{\mathrm{2}} }\:\:\:\:\:\Rightarrow\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\geqslant\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\: \\ $$$$\Leftrightarrow\mathrm{4z}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4x}^{\mathrm{2}} \right)\geqslant\mathrm{4x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4z}^{\mathrm{2}} \right)\Leftrightarrow\mathrm{z}^{\mathrm{2}} \geqslant\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{4x}^{\mathrm{2}} =\mathrm{x}\left(\mathrm{1}+\mathrm{4x}^{\mathrm{2}} \right)\Leftrightarrow\mathrm{4x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{x}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\right)=\mathrm{0}\Leftrightarrow\mathrm{x}\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{x}}\in\left\{\mathrm{0};\frac{\mathrm{1}}{\mathrm{2}}\right\}\:\mathrm{Hence},\mathrm{the}\:\mathrm{given}\:\mathrm{system} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{as} \\ $$$$\left\{\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)\in\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right);\left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\right. \\ $$
Commented by bemath last updated on 07/Aug/20
sir what is WLOG
$$\mathrm{sir}\:\mathrm{what}\:\mathrm{is}\:\mathrm{WLOG} \\ $$
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
$$\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{brief}\:\mathrm{writting}\:\mathrm{of}\:\mathrm{phrase} \\ $$$$“\mathrm{without}\:\mathrm{loss}\:\mathrm{the}\:\mathrm{generality}''\mathrm{is}\:\mathrm{used} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{math}.\:\mathrm{proof}\: \\ $$

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