## Subject II.

Let $x, y, z$ be real numbers such that $\frac{x y z}{x+y}=-1, \frac{x y z}{y+z}=1$ and $\frac{x y z}{z+x}=a$, where $a>\frac{1}{2}$ is a real number. Determine the product $x y z$.

Marin Chirciu, G.M. nr. 11/2012
## Solution.

From $\frac{x y z}{x+y}=-1$, it follows that $\frac{x+y}{x y z}=-1$, which means $\frac{1}{y z}+\frac{1}{x z}=-1$ (1)

Similarly, we obtain: $\frac{1}{x z}+\frac{1}{x y}=1$ (2) and $\frac{1}{x y}+\frac{1}{y z}=\frac{1}{a}$ (3) $\qquad$
Adding (1), (2), and (3) results in $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=\frac{1}{2 a}$

(4) $\qquad$
From (1) and (4), it follows that $\frac{1}{x y}=\frac{2 a+1}{2 a}$ (5) $\qquad$ (1p)

From (2) and (4), it follows that $\frac{1}{y z}=-\frac{2 a-1}{2 a}$ (6), and from (3) and (4) we have $\frac{1}{z x}=-\frac{1}{2 a}$ (7) $\qquad$ Multiplying relations (5), (6), and (7) yields $\left(\frac{1}{x y z}\right)^{2}=\frac{4 a^{2}-1}{8 a^{3}}$ and thus $x y z= \pm \frac{2 a \sqrt{2 a}}{\sqrt{4 a^{2}-1}} \ldots$