Observe that since $xy=1$, then $x+y \ge 2 \ge C$. Then, we see that $$\frac{C(x+y-2)}{2(x+y)} \le \frac{(x+y-2)}{2} $$must be true. If $C \le 0$ then $LHS \le 0 \le RHS$, if $C > 0$ then above inequality is equivalent to $x+y \ge C$ which is true. Rearranging that one, we see that $$1+\frac{C}{2} \le \frac{x+y}{2} + \frac{C}{x+y} \le \sqrt{\frac{x^2+y^2}{2}} + \frac{C}{x+y} $$by AM-QM.