To solve this problem, we need to find all real numbers \( a \) such that the system of equations has at least one real solution. The system of equations is: \[ x + \frac{a}{y} = y + \frac{a}{z} = z + \frac{a}{x}. \] ### Analysis Since each expression is equal, set them equal to a common value, say \( k \): \[ x + \frac{a}{y} = k, \] \[ y + \frac{a}{z} = k, \] \[ z + \frac{a}{x} = k. \] ### Solving the Equations From each equation, express \( x \), \( y \), and \( z \) in terms of \( k \) and \( a \): 1. \( x = k - \frac{a}{y} \) 2. \( y = k - \frac{a}{z} \) 3. \( z = k - \frac{a}{x} \) ### Symmetry and Substitution Notice the symmetry in the equations. Assume \( x = y = z \). Then: \[ x + \frac{a}{x} = k, \] which implies: \[ x = k - \frac{a}{x}. \] This leads to: \[ x^2 = xk - a. \] Rearrange this: \[ x^2 - xk + a = . \] ### Condition for Real Solutions For \( x \) to have real solutions, the discriminant of this quadratic equation must be non-negative: \[ k^2 - 4a \geq . \] Thus, \( a \leq \frac{k^2}{4} \). ### Conclusion Since \( k \) can be any real number, the condition \( a \leq \frac{k^2}{4} \) must hold for at least one \( k \). Therefore, the solution for \( a \) is: \[ a \leq . \] ### Final Answer All real numbers \( a \leq \) satisfy the condition where the system has at least one solution.