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 
 A: 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.
A: 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.  
A: 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 $\frac12$ .
