Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $\frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$. 
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $$\large \frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$$

Well, I have provided my solution below, which is clearly not the most straightforward and convenient one. I would be greatly appreciated if you could come up with any other solutions.

We have that $$\left\{ \begin{align} x + y &= m\\ xy &= n \end{align} \right. \implies m^2 - 3n - 1 = 0$$ and $$\frac{x}{x^2 + y} + \frac{y}{y^2 + x} = \frac{x(y^2 + x) + y(x^2 + y)}{(x^2 + y)(y^2 + x)} = \frac{mn + m^2 - 2n}{m^3 + n^2 - 3mn + n} = \frac{mn + n + 1}{n^2 + m + n}$$
It needs to be sufficient to prove that $$\frac{mn + n + 1}{n^2 + m + n} \le 1 \implies n^2 - mn + m - 1 \ge 0 \iff (xy)^2 - xy(x + y) + (x + y) + 1 \ge 0$$
$$\iff (xy - x - y + 1)(xy - 1) \ge 0 \implies (x - 1)(y - 1)(xy - 1) \ge 0$$
, which can be easily proven using Dirichlet's theorem.
