solve...

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?

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

but 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

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.)