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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

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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.

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.

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 $\frac12$ .

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