Let $y_i = 1/(x_i+1)$, $i = 1, 2, \dots, n$, so the $y_i$ are positive real numbers that add up to $1$. Upon substitution, the left-hand member of the required inequality becomes
$$\sum_{i=1}^{n} \frac{y_i^{\alpha}}{(1 - y_i)^{\alpha} + y_i^{\alpha}} = \sum_{i=1}^{n} \frac{y_i^{\alpha}}{\left(\sum_{j \neq i} y_j\right)^{\alpha} + y_i^{\alpha}},$$
the latter on account of $y_1 + y_2 + \dots + y_n = 1$. Apply Jensen's inequality to the convex function $t \mapsto t^{\alpha}$, $t > 0$, to get
$$\left( \sum_{j \neq i} y_j \right)^{\alpha} \le (n-1)^{\alpha-1} \sum_{j \neq i} y_j^{\alpha}, \quad i = 1, 2, \dots, n,$$
so
$$\sum_{i=1}^{n} \frac{y_i^{\alpha}}{\left(\sum_{j \neq i} y_j\right)^{\alpha} + y_i^{\alpha}} \ge \sum_{i=1}^{n} \frac{y_i^{\alpha}}{(n-1)^{\alpha-1} \sum_{j \neq i} y_j^{\alpha} + y_i^{\alpha}}.$$
Now write $z_i = y_i^{\alpha}$, $i = 1, 2, \dots, n$, $z = z_1 + z_2 + \dots + z_n$ and $a = (n-1)^{\alpha-1}$ to transform the right-hand member of the above inequality to
$$\sum_{i=1}^{n} \frac{z_i}{(1-a)z_i + az}.$$
Finally, notice that the function $t \mapsto \frac{t}{(1-a)t + az}$, $t < az/(a-1)$, is convex, to conclude by Jensen's inequality:
$$\sum_{i=1}^{n} \frac{z_i}{(1-a)z_i + az} \ge n \cdot \frac{\frac{1}{n} \sum_{i=1}^{n} z_i}{(1-a)^n \sum_{i=1}^{n} z_i + az} = \frac{n}{(n-1)a+1}.$$