solve...

$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 $x^3,y^3,z^3.$\\\\

@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

Its 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

So how can u make them equal?

xyz + z = xyz+ x = xyz + y
from these equations xyz cancels as xyz≠0

shubhodip
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

Suppose 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}

Suppose 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.)