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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