Solution: If $a=b=c=\frac{1}{3}$, then $\frac{(a+1)(b+1)(c+1)}{a b c+1}=\frac{16}{7}$. We will prove that $\frac{(a+1)(b+1)(c+1)}{a b c+1} \leq \frac{16}{7}$. Transform the expression to $\frac{(a+1)(b+1)(c+1)}{a b c+1}=\frac{a b c+a b+b c+a c+a+b+c+1}{a b c+1}=1+\frac{a b+b c+a c+1}{a b c+1}$. We will prove that $\frac{a b+b c+a c+1}{a b c+1} \leq \frac{9}{7}$. Make the substitution $c=1-a-b$ and we will prove the inequality

$$
7(a b+(a+b)(1-a-b)+1) \leq 9(a b(1-a-b)+1)
$$

which, after expanding the brackets and combining like terms, will take the form

$$
9 a^{2} b+9 a b^{2}-16 a b-7 a^{2}-7 b^{2}+7 a+7 b-2 \leq 0
$$

Since $a+b<1$, at least one of the variables is less than $\frac{7}{9}$—let this be $b$. Rewrite the inequality as

$$
(9 b-7) a^{2}+\left(9 b^{2}-16 b+7\right) a-7 b^{2}+7 b-2 \leq 0
$$

Since $9 b-7<0$, for the obtained inequality to hold for any $0<a<1$, it is sufficient to prove that the left side has no solutions with respect to $a$, that is, that the discriminant of the quadratic equation

$$
(9 b-7) a^{2}+\left(9 b^{2}-16 b+7\right) a-7 b^{2}+7 b-2=0
$$

is non-positive.

$$
\begin{gathered}
D=\left(9 b^{2}-16 b+7\right)^{2}+4(9 b-7)\left(7 b^{2}-7 b+2\right)=(9 b-7)^{2}(b-1)^{2}+4(9 b-7)\left(7 b^{2}-7 b+2\right) \\
=(9 b-7)\left(9 b^{3}+3 b^{2}-5 b+1\right)=(9 b-7)(b+1)(3 b-1)^{2} \leq 0
\end{gathered}
$$

which is what we needed to prove.

Answer: $\frac{16}{7}$
5. Positive numbers $a, b, c$ are such that $a+b+c=1$. Find the maximum value of the expression $\frac{(a+1)(b+1)(c+1)}{a b c+1} \cdot(20$ points $)$