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.
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 $\frac12$ .
$$(1-x)^3 \le (\sqrt{2} - 1)^3$$ $$1-x \le \sqrt{2} -1$$ $$2-x \le \sqrt{2}$$ $$x \ge 2 - sqrt{2}$$
So this is only satisfied for $x \ge 0.59$, but x can be smaller than 0.59, so this won't work. One good appraoch is that maximum of the RHS is 3, so try somehow to prove that LHS is bigger than 3.
– Stefan4024 Sep 29 '13 at 18:32