Let $a,b,c$ be a real numbers such that this equations: $a^2x + b^2y + c^2z = 1$ $xy + yz + xz = 1$ have only one solution $(x, y, z)$ in real numbers. Prove that $a, b, c$ are sides of the triangle
Source: Peru TST ibero
Tags: algebra
Let $a,b,c$ be a real numbers such that this equations: $a^2x + b^2y + c^2z = 1$ $xy + yz + xz = 1$ have only one solution $(x, y, z)$ in real numbers. Prove that $a, b, c$ are sides of the triangle