37. Given that $a, b, c$ are non-negative real numbers, and $a+b+c \leqslant 3$, find the minimum value of $\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+$ $\frac{c+1}{c(c+2)}$. (2003 Austrian Mathematical Olympiad Problem)
37. Let $x=a+1, y=b+1, z=c+1$, then the given conditions become $x, y, z \geqslant 1, x+y+z \leqslant 6$, and the problem becomes finding the minimum value of $\frac{x}{x^{2}-1}+\frac{y}{y^{2}-1}+\frac{z}{z^{2}-1}$.

When $x \geqslant 1$, it is easy to prove that $\frac{x}{x^{2}-1} \geqslant \frac{16-5 x}{9}$.
In fact,
$$\frac{x}{x^{2}-1} \geqslant \frac{16-5 x}{9}(x-1)^{2}(5 x+4) \geqslant 0$$

Similarly, $\frac{y}{y^{2}-1} \geqslant \frac{16-5 y}{9}, \frac{z}{z^{2}-1} \geqslant \frac{16-5 z}{9}$. Adding these, and noting that $x+y+z \leqslant 6$, we get
$$\frac{x}{x^{2}-1}+\frac{y}{y^{2}-1}+\frac{z}{z^{2}-1} \geqslant \frac{48-5(x+y+z)}{9} \geqslant \frac{48-30}{9}=2$$

Therefore, the minimum value of $\frac{x}{x^{2}-1}+\frac{y}{y^{2}-1}+\frac{z}{z^{2}-1}$ is 2, which occurs if and only if $x=y=z=2$, i.e., if and only if $a=b=c=1$, the minimum value of $\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)}$ is 2.