Solution. Notice first that it is sufficient to prove the inequality in the case $n=3$. For a bigger value of $n(n \geq 4)$, we only need to choose from the set $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ the three smallest numbers, say $a_{1}, a_{2}, a_{3}$. Since $a_{1} a_{2} a_{3} \leq 1$, there exists a positive number $k$ such that $\frac{a}{a_{1}}=\frac{b}{a_{2}}=\frac{c}{a_{3}}=k \geq 1$, then
$$\sum_{i=1}^{n} \frac{a_{i}+3}{\left(a_{i}+1\right)^{2}} \geq \frac{a_{1}+3}{\left(a_{1}+1\right)^{2}}+\frac{a_{2}+3}{\left(a_{2}+1\right)^{2}}+\frac{a_{3}+3}{\left(a_{3}+1\right)^{2}} \geq \sum_{c y c} \frac{a+3}{(a+1)^{2}} \geq 3$$

We will now prove that if $a, b, c$ are positive real numbers and $a b c=1$ then
$$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}} \geq 3$$

Let $a_{1}=\frac{2}{1+a}, b_{1}=\frac{2}{1+b}, c_{1}=\frac{2}{1+c}$. The inequality becomes
$$a_{1}+b_{1}+c_{1}+a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \geq 6$$

Since $a b c=1$, we have
$$\prod_{c y c}\left(\frac{1}{a_{1}}-\frac{1}{2}\right)=\frac{a b c}{8}=\frac{1}{8} \Rightarrow \prod_{c y c}\left(2-a_{1}\right)=a_{1} b_{1} c_{1}$$

Let $x=a_{1}-1, y=b_{1}-1, z=c_{1}-1$, then $x, y, z \in[-1,1]$ and we infer that
$$(x+1)(y+1)(z+1)=(1-x)(1-y)(1-z) \Rightarrow x+y+z+x y z=0$$

By AM-GM inequality, we deduce that $x^{2}+y^{2}+z^{2} \geq 3(x y z)^{2 / 3} \geq 3 x y z$, thus
$$a_{1}+b_{1}+c_{1}+a_{1}^{2}+b_{1}^{2}+c_{1}^{2}-6=\sum_{c y c}\left(a_{1}-1\right)\left(a_{1}+2\right)=\sum_{c y c} x(x+3) \geq 0$$

This ends the proof. Equality holds for $a=b=c=1$.