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 

The only $(x,y,z)$ is $(\frac12,\frac12,\frac12)$.
For all other $x,y,z \in (0,1)$ the left-hand side is larger.
We prove this by showing that if we fix the average $a=\frac13(x+y+z)$ then
(i) the left-hand side is minimized when $x=y=z=a$, while
(ii) the right-hand side is maximized when $x=y=z=a$.
Hence it is enough to show the claimed inequality when $x=y=z=a$, which is
$$
\left(a + \frac1{2a} - 1\right)^3 \geq (1-a)^3
$$
with equality iff $a=1/2$.  Equivalently, we claim
$a + \frac1{2a} - 1 \geq 1-a$ with the same equality condition;
and this easy because the difference between the sides is $(2a-1)^2/(2a)$.
It remains to prove assertions (i) and (ii).  For (i),
the OP already noted that each factor is bounded below by $\sqrt 2 - 1$,
so in particular the factors are all positive.
We compute that $\log(x + \frac1{2x} - 1)$ is convex upwards by calculating
$$
\frac{d^2}{dx^2} \log\left(x + \frac1{2x} - 1\right)
= \frac{1-4x+8x^2-4x^4}{x^2(1-2x+2x^2)^2}
$$
and showing that the numerator is positive for $0<x<1$:
$$
1 - 4x + 8x^2 - 4x^4 > 1 - 4x + 8x^2 - 4x^3
= 1 - 4x(1-x)^2 > 1-4x(1-x) = (2x-1)^2.
$$
For (ii), we may assume that each of the factors is positive:
at most one of $xy/z$, $yz/x$, and $xz/y$ can be $\geq 1$
(if two of them are, then so is their product, which is
$x^2$, $y^2$, or $z^2$, contradicting $x,y,z \in (0,1)$);
and if exactly one factor is not positive, then
the right-hand side is $\leq 0$ and we're done.
Once all three factors are positive, we have
(because $\log(1-x)$ is concave downwards)
$$
\left( 1 - \frac{xy}{z} \right)
\left( 1 - \frac{yz}{x} \right)
\left( 1 - \frac{xz}{y} \right)
\leq (1-c)^3
$$
where $c$ is the average of $xy/z$, $yz/x$, and $xz/y$.
But $c \geq a$ with equality iff $x=y=z=a$: we have
$\frac12(\frac{xy}{z} + \frac{yz}{x}) \geq y$
by the inequality on arithmetic and geometric means, and likewise
$\frac12(\frac{yz}{x} + \frac{zx}{y}) \geq z$ and
$\frac12(\frac{zx}{y} + \frac{xy}{z}) \geq x$;
and summing these three inequalities yields
$$
 \frac{xy}{z} + \frac{yz}{x} + \frac{xz}{y} \geq x + y + z  = 3a
$$
as claimed.  Hence $(1-c)^3 \leq (1-a)^3$ and we're done.
