MS_Kekas wrote: Determine the largest possible constant \( C \) such that for any positive real numbers \( x, y, z \), which are the sides of a triangle, the following inequality holds: \[ \frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C. \] Proposed by Vadym Solomka WLOG that \( y = \max\{x, y, z\} \). When \( x = y \) and \( z \to 0 \), \( \mathrm{LHS} \to \frac{1}{2} \), thus \( C \leq \frac{1}{2} \). On the other hand, we have \[ y^2 + z^2 + yx < y^2 + y^2 + (x + z)x < 2(x^2 + y^2 + xz). \] Therefore, \[ \frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} > \frac{xy}{x^2 + y^2 + xz} + \frac{yz}{2(x^2 + y^2 + xz)} = \frac{2xy + yz}{2(x^2 + y^2 + xz)}. \] Moreover, \[ 2xy + yz - (x^2 + y^2 + xz) = (x + z - y)(y - x) \geq 0, \] thus \[ \frac{2xy + yz}{2(x^2 + y^2 + xz)} \geq \frac{1}{2}. \] This demonstrates that \( C = \frac{1}{2} \) can be achieved, hence \( C_{\text{max}} = \frac{1}{2} \).