To find the largest positive real number $p$ such that the inequality
$$x_1^2 + x_2^2 + \cdots + x_n^2 \ge p(x_1x_2 + x_2x_3 + \cdots + x_{n-1}x_n)$$
is satisfied for all real numbers $x_i$, we will consider the cases $n = 2$, $n = 5$, and general $n \ge 2$.

### Case (a): $n = 2$

1. The inequality becomes:
$$x_1^2 + x_2^2 \ge p x_1 x_2.$$

2. Using the AM-GM inequality:
$$x_1^2 + x_2^2 \ge 2x_1x_2.$$

3. Comparing this with the given inequality, we see that:
$$2x_1x_2 \ge p x_1 x_2.$$

4. For the inequality to hold for all $x_1$ and $x_2$, we must have:
$$2 \ge p.$$

Thus, the largest value of $p$ for $n = 2$ is $p = 2$.

### Case (b): $n = 5$

1. The inequality becomes:
$$x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 \ge p(x_1x_2 + x_2x_3 + x_3x_4 + x_4x_5).$$

2. Using the AM-GM inequality in a more detailed manner:
$$\left( x_1^2 + \frac{x_2^2}{3} \right) + \left( \frac{2x_2^2}{3} + \frac{x_3^2}{2} \right) + \left( \frac{x_3^2}{2} + \frac{2x_4^2}{3} \right) + \left( \frac{x_4^2}{3} + x_5^2 \right) \ge \frac{2}{\sqrt{3}}(x_1x_2 + x_2x_3 + x_3x_4 + x_4x_5).$$

3. Simplifying the left-hand side:
$$x_1^2 + \frac{x_2^2}{3} + \frac{2x_2^2}{3} + \frac{x_3^2}{2} + \frac{x_3^2}{2} + \frac{2x_4^2}{3} + \frac{x_4^2}{3} + x_5^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2.$$

4. Therefore, we have:
$$x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 \ge \frac{2}{\sqrt{3}}(x_1x_2 + x_2x_3 + x_3x_4 + x_4x_5).$$

5. Comparing this with the given inequality, we see that:
$$p \le \frac{2}{\sqrt{3}}.$$

Thus, the largest value of $p$ for $n = 5$ is $p = \frac{2}{\sqrt{3}}$.

### General Case: $n \ge 2$

1. We need to find the largest $p$ such that:
$$x_1^2 + x_2^2 + \cdots + x_n^2 \ge p(x_1x_2 + x_2x_3 + \cdots + x_{n-1}x_n).$$

2. Using the Cauchy-Schwarz inequality:
$$(x_1^2 + x_2^2 + \cdots + x_n^2)(1^2 + 1^2 + \cdots + 1^2) \ge (x_1 + x_2 + \cdots + x_n)^2.$$

3. Simplifying, we get:
$$n(x_1^2 + x_2^2 + \cdots + x_n^2) \ge (x_1 + x_2 + \cdots + x_n)^2.$$

4. For the inequality to hold for all $x_i$, we need:
$$x_1^2 + x_2^2 + \cdots + x_n^2 \ge \frac{(x_1 + x_2 + \cdots + x_n)^2}{n}.$$

5. Using the AM-GM inequality, we have:
$$x_1^2 + x_2^2 + \cdots + x_n^2 \ge \frac{(x_1 + x_2 + \cdots + x_n)^2}{n}.$$

6. Therefore, the largest $p$ for general $n$ is:
$$p = 2.$$

The final answer is $\boxed{ p = 2 }$ for $n = 2$, $p = \frac{2}{\sqrt{3}}$ for $n = 5$, and $p = 2$ for general $n \ge 2$.