1. Define the sequence and the sum:
Let $(x_n)_{n=1}^\infty$ be a sequence defined recursively by:
$$x_1 = 2 \quad \text{and} \quad x_{n+1} = \frac{x_n(x_n + n)}{n+1} \quad \text{for all} \ n \geq 1.$$
Define $S_n = x_1 + x_2 + \ldots + x_n$.

2. State the inequality to prove:
We need to prove that:
$$n(n+1) > \frac{(x_1 + x_2 + \ldots + x_n)^2}{x_{n+1}}.$$
This can be rewritten using $S_n$ as:
$$S_n < \sqrt{n(n+1)x_{n+1}}.$$

3. Base case:
For $n = 1$:
$$S_1 = x_1 = 2 \quad \text{and} \quad x_2 = \frac{x_1(x_1 + 1)}{2} = \frac{2 \cdot 3}{2} = 3.$$
We need to check:
$$2 < \sqrt{1 \cdot 2 \cdot 3} = \sqrt{6}.$$
Since $2 < \sqrt{6}$, the base case holds.

4. Inductive step:
Assume the inequality holds for some $n \geq 1$, i.e.,
$$S_n < \sqrt{n(n+1)x_{n+1}}.$$
We need to show that:
$$S_{n+1} < \sqrt{(n+1)(n+2)x_{n+2}}.$$
Note that:
$$S_{n+1} = S_n + x_{n+1}.$$
By the inductive hypothesis:
$$S_{n+1} = S_n + x_{n+1} < \sqrt{n(n+1)x_{n+1}} + x_{n+1}.$$

5. Simplify the inequality:
We need to prove:
$$\sqrt{n(n+1)x_{n+1}} + x_{n+1} < \sqrt{(n+1)(n+2)x_{n+2}}.$$
Using the recursive definition:
$$(n+2)x_{n+2} = x_{n+1}(x_{n+1} + n + 1).$$
Thus, we need to show:
$$\sqrt{n(n+1)x_{n+1}} + x_{n+1} < \sqrt{(n+1)x_{n+1}(x_{n+1} + n + 1)}.$$

6. Algebraic manipulation:
Square both sides to eliminate the square roots:
$$(\sqrt{n(n+1)x_{n+1}} + x_{n+1})^2 < (n+1)x_{n+1}(x_{n+1} + n + 1).$$
Expanding both sides:
$$n(n+1)x_{n+1} + 2x_{n+1}\sqrt{n(n+1)x_{n+1}} + x_{n+1}^2 < (n+1)x_{n+1}^2 + (n+1)x_{n+1}(n+1).$$
Simplify:
$$n(n+1)x_{n+1} + 2x_{n+1}\sqrt{n(n+1)x_{n+1}} + x_{n+1}^2 < (n+1)x_{n+1}^2 + (n+1)^2x_{n+1}.$$
Combine like terms:
$$2x_{n+1}\sqrt{n(n+1)x_{n+1}} < (n+1)^2x_{n+1} - n(n+1)x_{n+1}.$$
Factor out $x_{n+1}$:
$$2\sqrt{n(n+1)x_{n+1}} < (n+1)x_{n+1} + (n+1)x_{n+1}.$$
Simplify:
$$2\sqrt{n(n+1)x_{n+1}} < (2n+1)x_{n+1}.$$
Divide both sides by $x_{n+1}$:
$$2\sqrt{n(n+1)} < 2n+1.$$
This is immediate from the AM-GM inequality and the fact that $2n+1 > n+1$.

Thus, the inequality holds for all $n \geq 1$ by induction.

$\blacksquare$