China Mathematical Competition

$|\lg(a+1)| = |\lg\left(-\frac{b+1}{b+2}+1\right)| = |\lg\left(\frac{1}{b+2}\right)| = |\lg(b+2)|.$
Then either $a+1 = b+2$ or $(a+1)(b+2) = 1$. Since $a < b$, so $a+1 \neq b+2$. Therefore, $(a+1)(b+2) = 1$.
From $f(a) = |\lg(a+1)|$ we know $0 < a+1 < 1$. Then
$$0 < a + 1 < b + 1 < b + 2,$$
which implies
$$0 < a + 1 < 1 < b + 2.$$
Therefore,
$$(10a + 6b + 21) + 1 = 10(a + 1) + 6(b + 2) \\ = 6(b + 2) + \frac{10}{b + 2} > 1.$$
Then
$$f(10a + 6b + 21) = \left| \lg \left[ 6(b + 2) + \frac{10}{b+2} \right] \right| \\ = \lg \left[ 6(b + 2) + \frac{10}{b+2} \right].$$
On the other hand,
$$f(10a + 6b + 21) = 4\lg 2.$$
So
$$\lg\left[6(b+2) + \frac{10}{b+2}\right] = 4\lg 2,$$
which means $6(b+2) + \frac{10}{b+2} = 16$. Then either $b = -\frac{1}{3}$ or $b = -\frac{2}{5}$ (discarded).
Substituting $b = -\frac{1}{3}$ into $(a+1)(b+2) = 1$, we find $a = -\frac{2}{5}$.
Therefore $a = -\frac{2}{5}$, $b = -\frac{1}{3}$.
$\square$