36So how can u make them equal?
If x + 1/y = y + 1/z = z + 1/x then prove that, xyz = x3 = y3 = z3 in simple way... please someone..!
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1708$Here $x+\frac{1}{x}=y+\frac{1}{y}=z+\frac{1}{z}$\\\\\ $x+\frac{1}{x}=y+\frac{1}{y}\Leftrightarrow x-y=\frac{1}{y}-\frac{1}{x}\Leftrightarrow (x-y)=\frac{(x-y)}{xy}$\\\\ $(x-y).(1-\frac{1}{xy})=0\Leftrightarrow \boxed{x=y}$ OR $\boxed{xy=1\Leftrightarrow xyz=z}..........................(1)$\\\\ Similarly\\\\ $y+\frac{1}{y}=z+\frac{1}{z}\Leftrightarrow y-z=\frac{1}{z}-\frac{1}{y}\Leftrightarrow (y-z)=\frac{(y-z)}{yz}$\\\\ $(y-z).(1-\frac{1}{yz})=0\Leftrightarrow \boxed{y=z}$ OR $yz=1\Leftrightarrow \boxed{xyz=x}............................(2)$\\\\ Similarly\\\\ $z+\frac{1}{z}=x+\frac{1}{x}\Leftrightarrow z-x=\frac{1}{x}-\frac{1}{z}\Leftrightarrow (z-x)=\frac{(z-x)}{zx}$\\\\ $(z-x).(1-\frac{1}{zx})=0\Leftrightarrow \boxed{z=x}$ OR $zx=1\Leftrightarrow \boxed{xyz=y}.............................(3)$\\\\ So from Equation....(1),(2) and (3), We Get,\\\\ $\boxed{\boxed{x=y=z=xyz}}$ \\\\ Now We will use $\underline{\underline{A.M\geq G.M}}$ for $x^3,y^3,z^3.$\\\\
36@man111 ->
What have u done?
The question says something else.....!
its x + 1/y = y + 1/z = z + 1/x and not, x + 1/x = y + 1/y = z + 1/z
1Its really simple
see
x + 1y = y + 1z = z + 1x
≡ xy+1y = yz+1z =xz+1x
cross multiplying every term with one another
xyz + z = xyz+ x = xyz + y
from which we obtain
x= y =z
and so
xyz = x3 = y3 = z3
1xyz + z = xyz+ x = xyz + y
from these equations xyz cancels as xyz≠0
1shubhodip
xyz ≠0 as x≠0 y≠0 z≠0 as 1/x is only applicable if x≠0
if xy or yz or zx equal to -1 then
xyz + z = xyz+ x = xyz + y = 0
then
x= y =z = ± 1
which gives the result
21but it must be given that they are positive
here is my soltuion
xyz= y2z+y-z
xyz= z2x+z-x
xyz = x2y+ x-y
adding all of them we get
3xyz = x2y + z2x + y2z
from AM>= GM we always have x2y + z2x + y2z >= 3xyz
equality holds when terms are equal
so x2y= z2x
as x not equal to zero we have xyz= z3
but to apply AM-GM terms must be positive,so it must be given x,y,z>0
341Suppose x≠y≠z
x + \frac{1}{y} = y + \frac{1}{z} \Rightarrow yz = \frac{y-z}{x-y}
Similarly,
xz = \frac{x-z}{y-z}; xy = \frac{x-y}{x-z}
Multiplying we obtainx^2y^2z^2=1 \Rightarrow xyz= \pm 1
If xyz =1, then there exist reals p,q,r such that x=\frac{p}{q}, y = \frac{q}{r}; z = \frac{r}{p}
341Suppose x≠y≠z
x + \frac{1}{y} = y + \frac{1}{z} \Rightarrow yz = \frac{y-z}{x-y}
Similarly,
xz = \frac{x-z}{y-z}; xy = \frac{x-y}{x-z}
Multiplying we obtain x^2y^2z^2=1 \Rightarrow xyz= \pm 1
If xyz =1, then there exist numbers p,q,r such that
x=\frac{p}{q}, y = \frac{q}{r}; z = \frac{r}{p}
Then plugging these back in the equation, we have
\frac{p+q}{r} = \frac{q+r}{p} = \frac{r+p}{q} = \frac{2(p+q+r)}{p+q+r}=2
and so
p+q=2r; q+r=2p; r+p=2q \Rightarrow p=q=r \Rightarrow x=y=z=1
(If p+q+r=0, we again get x=y=z, you can try and prove this)
Now, if xyz=-1, we have x = \frac{p}{q}; y = \frac{q}{r}; z = -\frac{r}{p}
Now, we have \frac{p+r}{q} = \frac{q-p}{r} = \frac{q-r}{p}
From the 1st two terms, this ratio =1.
Hence we have for any p,q,r such that (p+q)(q+r)(r+p)≠0,
and p+r=1, x=p/q, y=q/r, z=-r/p as a solution. (but we donot have xyz=x3 etc.)