Given a prime $p$, let $d(a,b)$ be the number of integers $c$ such that $1 \leq c < p$, and the remainders when $ac$ and $bc$ are divided by $p$ are both at most $\frac{p}{3}$. Determine the maximum value of \[\sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)(x_a + 1)(x_b + 1)} - \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)x_ax_b}\] over all $(p-1)$-tuples $(x_1,x_2,\ldots,x_{p-1})$ of real numbers.

[i]Brian Hamrick.[/i]
1. **Define the function \( d(a, b) \):**
   Given a prime \( p \), let \( d(a, b) \) be the number of integers \( c \) such that \( 1 \leq c < p \), and the remainders when \( ac \) and \( bc \) are divided by \( p \) are both at most \( \frac{p}{3} \).

2. **Express the given expression in terms of vectors:**
   Let \( v_a \) be a vector in \( \mathbb{R}^{p-1} \) such that the \( c \)-th term of \( v_a \) is:
   \[
   v_a(c) = 
   \begin{cases} 
   x_a + 1 & \text{if the remainder of } ac \text{ is at most } \frac{p}{3}, \\
   0 & \text{otherwise}.
   \end{cases}
   \]
   Similarly, let \( u_a \) be a vector in \( \mathbb{R}^{p-1} \) such that the \( c \)-th term of \( u_a \) is:
   \[
   u_a(c) = 
   \begin{cases} 
   x_a & \text{if the remainder of } ac \text{ is at most } \frac{p}{3}, \\
   0 & \text{otherwise}.
   \end{cases}
   \]

3. **Rewrite the sums using these vectors:**
   The given expression can be rewritten as:
   \[
   \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1} d(a, b) (x_a + 1)(x_b + 1)} = \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle}
   \]
   and
   \[
   \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1} d(a, b) x_a x_b} = \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle}.
   \]

4. **Apply the triangle inequality:**
   By the triangle inequality for vectors, we have:
   \[
   \| v_a \| \leq \| u_a \| + \| \mathbf{1} \|,
   \]
   where \( \mathbf{1} \) is the vector with all entries equal to 1. Since \( \| \mathbf{1} \| = \sqrt{p-1} \), we get:
   \[
   \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle} \leq \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle} + \sqrt{p-1}.
   \]

5. **Determine the maximum value:**
   The maximum value of the given expression is achieved when the difference between the two square roots is maximized. This occurs when:
   \[
   \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle} - \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle} = \sqrt{p-1}.
   \]

The final answer is \(\boxed{\sqrt{p-1}}\).