Find all triples $(x,y,z)$ of real numbers with $x,y,z>1$ satisfying
$$\left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)=\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right).$$
Proposed by Alessandro Ventullo, Milan, Italy
Solution: By the AM-GM Inequality, we have
$$\dfrac{x}{2}+\dfrac{1}{x}-1=\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)+\dfrac{1}{2x}-1 \geq \dfrac{1}{2x} > 0$$
$$\dfrac{y}{2}+\dfrac{1}{y}-1=\left(\dfrac{y}{2}+\dfrac{1}{2y}\right)+\dfrac{1}{2y}-1 \geq \dfrac{1}{2y} > 0$$
$$\dfrac{z}{2}+\dfrac{1}{z}-1=\left(\dfrac{z}{2}+\dfrac{1}{2z}\right)+\dfrac{1}{2z}-1 \geq \dfrac{1}{2z} > 0.$$
Thus, $$\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)>0.$$
Assume that two among these three factors are negative. Without loss of generality, $1-\dfrac{x}{yz}<0$ and $1-\dfrac{y}{zx}<0$. Then, $\dfrac{x}{yz}>1$ and $\dfrac{y}{zx}>1$, which gives $\dfrac{1}{z^2}>1$, contradiction. Hence,
$$1-\dfrac{x}{yz}>0, \qquad 1-\dfrac{y}{zx}>0, \qquad 1-\dfrac{z}{xy}>0.$$
We have $$\begin{array}{lll} \left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)^2-\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)&=&\dfrac{1}{x^2}+\dfrac{2}{x}\left(\dfrac{x}{2}-1\right)+\left(\dfrac{x}{2}-1\right)^2-\left[1-\dfrac{yz}{x}\left(\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)+\dfrac{1}{x^2}\right] \\ &=&\left(\dfrac{x}{2}-1\right)^2+\dfrac{yz}{x}\left(\dfrac{1}{y}-\dfrac{1}{z}\right)^2 \geq 0. \end{array}$$
The equality holds if and only if $x=2$ and $y=z$. Likewise,
$$\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)^2 \geq \left(1-\dfrac{z}{xy}\right)\left(1-\dfrac{x}{yz}\right)$$
$$\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)^2 \geq \left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)$$
Multiplying these three inequalities, we get that the left-hand side of the given equation is greater or equal than the right-hand side and the equality holds if and only if $x=y=z=2$. So, $(x,y,z)=(2,2,2)$.
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