Find the largest positive real number $p$ (if it exists) such that the inequality
\[x^2_1+ x_2^2+ \cdots + x^2_n\ge p(x_1x_2 + x_2x_3 + \cdots + x_{n-1}x_n)\]
is satisfied for all real numbers $x_i$, and $(a) n = 2; (b) n = 5.$
Find the largest positive real number $p$ (if it exists) such that the inequality holds for all real numbers $x_i$ and all natural numbers $n, n \ge 2.$
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 \( n \ge 2 \).