Question 73: Let $x \geq 1, y \geq 1$, for any real number $k$ there is $\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+(x-1)(y-1) \geq k x y$, find the maximum value of $k$.


Question 73, Solution: On one hand, when $x=y=\frac{1+\sqrt{5}}{2}$, we have $2+\left(\frac{\sqrt{5}-1}{2}\right)^{2} \geq k\left(\frac{1+\sqrt{5}}{2}\right)^{2}$, solving this gives $k \leq \frac{13-5 \sqrt{5}}{2}$.

On the other hand, we prove that for any $x, y \in [1, +\infty)$, it always holds that:
$$
\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+(x-1)(y-1) \geq \frac{13-5 \sqrt{5}}{2} x y
$$
In fact, transforming the inequality, we get:
$$
\begin{array}{l}
\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+(x-1)(y-1) \geq \frac{13-5 \sqrt{5}}{2} x y \\
\Leftrightarrow \frac{5 \sqrt{5}-11}{2} x y + 1 \geq \left(x - \frac{x^{2}}{1+x}\right) + \left(y - \frac{y^{2}}{1+y}\right) = \frac{x}{1+x} + \frac{y}{1+y} \\
\Leftrightarrow \frac{5 \sqrt{5}-11}{2} \geq \frac{x y - 1}{x y (1+x)(1+y)}
\end{array}
$$

In fact, let $u = \sqrt{x y} - 1$, then $u \geq 0$. According to the Cauchy-Schwarz inequality, we have:
$$
\frac{x y - 1}{x y (1+x)(1+y)} \leq \frac{x y - 1}{x y (1+\sqrt{x y})^{2}} = \frac{\sqrt{x y} - 1}{x y (1+\sqrt{x y})} = \frac{u}{(u+1)^{2}(u+2)}
$$

Thus, we only need to prove:
$$
\frac{5 \sqrt{5}-11}{2} \geq \frac{u}{(u+1)^{2}(u+2)} \Leftrightarrow \left(u - \frac{\sqrt{5}-1}{2}\right)^{2} \geq 0
$$

The above inequality is clearly true.
In summary, the maximum value of $k$ is $\frac{13-5 \sqrt{5}}{2}$.