For the height 553, find the largest real number $M$ such that for all real numbers $x, y$ satisfying $x+y \geqslant 0$, we have
$$
\left(x^{2}+y^{2}\right)^{3} \geqslant M\left(x^{3}+y^{3}\right)(x y-x-y).
$$
The maximum value of the required $M$ is 32.
First, take $x=y=4$, we get $M \leqslant 32$.
Next, we prove: for all real numbers $x, y$ satisfying $x+y \geqslant 0$, we have
$$
\begin{array}{l}
\left(x^{2}+y^{2}\right)^{3} \geqslant 32\left(x^{3}+y^{3}\right)(x y-x-y) . \\
\text { Let } s=x^{2}+y^{2}, t=x+y(t \geqslant 0) .
\end{array}
$$

By the AM-GM inequality, we know $2 s \geqslant t^{2}$.
Thus, the inequality to be proven is transformed into
$$
s^{3} \geqslant 8 t\left(3 s-t^{2}\right)\left(t^{2}-2 t-s\right) \text {. }
$$

Let $s=r t$.
Then equation (1) is equivalent to
$$
r^{3} \geqslant 8(3 r-t)(t-2-r)(2 r \geqslant t \geqslant 0) \text {. }
$$

In fact,
$$
\begin{array}{l}
r^{3}-8(3 r-t)(t-2-r) \\
=8(t-2 r-1)^{2}+r^{3}-8 r^{2}+16 r-8 \\
\geqslant r^{3}-8 r^{2}+16 r=r(r-4)^{2} \geqslant 0 .
\end{array}
$$

Clearly, equation (1) holds.