How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)$$
my try:use
$$x+\dfrac{1}{2x}-1\ge 2\sqrt{\dfrac{1}{2}}-1=\sqrt{2}-1$$
so
$$LHS\ge (\sqrt{2}-1)^3$$
so I guess 
$$RHS\le (\sqrt{2}-1)^3$$
But I can't prove it.Thank you 

My approach lacks something at the end.
$$x+\dfrac{1}{2x}-1 = \frac{2x^2}{2x}-\frac{2x}{2x}+\dfrac{1}{2x} = \frac{2x^2 -2x +1}{2x}$$
$$1 - \frac{xy}{z} = \frac{z-xy}{z}$$
$$
\begin{align}
\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)&=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)\\
\frac{2x^2 -2x +1}{2x}\frac{2y^2 -2y +1}{2y}\frac{2z^2 -2z +1}{2z} &= \frac{z-xy}{z}\frac{x-yz}{x}\frac{y-zx}{y}\\
(2x^2 -2x +1)(2y^2 -2y +1)(2z^2 -2z +1) &= 8(z-xy)(x-yz)(y-zx)\\
\end{align}
$$
Clearly,
$$\dfrac{1}{8} \leq (2x^2 -2x +1)(2y^2 -2y +1)(2z^2 -2z +1) \lt 1$$
My work in progress is to show that $(z-xy)(x-yz)(y-zx) \leq \dfrac{1}{64}$, but I thought I'd put this up as it seems a simpler proof for the LHS.  
