We first observe that $x^4+3 \ge (x+1)^2$ as $x^4+3-(x+1)^2 = (x-1)^2(x^2+2x+2) \ge 0$. Therefore it suffices to prove that
$$\sum_{i=1}^{n} \frac{a_i}{a_i + 1} \le \frac{1}{2} \sum_{i=1}^{n} \frac{1}{a_i}$$
for all positive integers $n$ and positive real numbers $a_1, a_2, \dots, a_n$ satisfying the condition $a_1 a_2 \cdots a_n = 1$.

Next we observe that
$$\frac{1}{2xy} - \frac{xy}{xy+1} \le \frac{1}{2x} + \frac{1}{2y} - \frac{x}{x+1} - \frac{y}{y+1} \quad (*)$$
for $0 < x \le 1 \le y$. This can be seen using the fact that
$$0 \le (1-x)(y-1)(2x^2y^2 + x^2y + xy^2 + xy + x + y + 1)$$
after collecting all the terms to the right side of the inequality and getting a common denominator.

Let $f_k(x_1, x_2, \dots, x_k) = \frac{1}{2} \sum_{i=1}^{k} \frac{1}{x_i} - \sum_{i=1}^{k} \frac{x_i}{x_i + 1}$. We will prove by induction on $k$ that $x_1 x_2 \cdots x_k = 1 \implies f_k(x_1, x_2, \dots, x_k) \ge 0$.

* For $k=1$, $x_1=1$ and $f_1(x_1) = 0$.

* Suppose $k > 1$ and $x_1 x_2 \cdots x_k = 1$. If $x_1 \le x_2 \le \cdots \le x_k$, then $x_1 \le 1 \le x_k$, and using (*) we get $f_k(x_1, x_2, \dots, x_k) \ge f_{k-1}(x_1 x_k, x_2, \dots, x_{k-1}) \ge 0$.