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 

It is fairly obvious that, since the expressions on both sides are symmetrical with regards to each of the three variables, they can easily be reduced to $$\left(t + \frac1{2t} - 1\right)^3 = \left(1 - \frac{t\ \cdot\ t}t\right)^3 \qquad\iff\qquad \left(t + \frac1{2t} - 1\right) = (1 - t) \quad | \cdot 2t$$ $$2t^2 + 1 - 2t\ =\ 2t - 2t^2 \qquad\iff\qquad 4t^2 - 4t + 1\ =\ 0 \quad\iff\quad (2t - 1)^2\ =\ 0$$ whose only solution is $t = \frac12$ , which is the only point of intersection between the hyperbola on the left, and the linear polynomial on the right.

Another approach, also exploiting symmetry: Let $$f_t(u,v) = t + \frac1{2t} - 1 \qquad ; \qquad g_t(u,v) = 1 - \frac{u\ \cdot\ v}t$$ Then our equation becomes $$f_x(y,z) \cdot f_y(x,z) \cdot f_z(x,y)\ =\ g_x(y,z) \cdot g_y(x,z) \cdot g_z(x,y)$$ which, due to its inherent permutability, is ultimately reduced to $f_t(u,v) = g_t(u,v)$ , for all three possible cases. This, in its turn, yields $$t + \frac1{2t} - 1 = 1 - \frac{uv}t\ \iff\ 2t^2 + 1 - 2t = 2t - 2uv\ \iff\ 2t^2 - 4t + (2uv + 1) = 0$$ from where we get the formula $t = 1 - \sqrt{\frac12 - uv}$ , which, when spelled out for all three variables, becomes $$x = 1 - \sqrt{\tfrac12 - yz} \qquad,\qquad y = 1 - \sqrt{\tfrac12 - xz} \qquad,\qquad z = 1 - \sqrt{\tfrac12 - xy}$$ By extracting $x$ from the latter two in terms of $y$ and $z$, we get $$x = \frac{1 - 2(1-y)^2}z \qquad,\qquad x = \frac{1 - 2(1-z)^2}y$$ $$\iff\quad y\cdot[1 - 2(1-y)^2]\ =\ z\cdot[1 - 2(1-z)^2] \quad\iff\quad y = z$$ Analogously, we show that $x = y$ and $x = z$ , ultimately leading to $x = y = z$ , for which the only possible solution is $\frac12$ .
