Given $n \geq 2$ reals $x_1 , x_2 , \dots , x_n.$ Show that
$$\prod_{1\leq i < j \leq n} (x_i - x_j)^2 \leq \prod_{i=0}^{n-1} \left(\sum_{j=1}^{n} x_j^{2i}\right)$$
and find all the $(x_1 , x_2 , \dots , x_n)$ where the equality holds.
1. **Hadamard's Inequality for Vandermonde Matrix**:
   The given inequality can be approached using Hadamard's inequality for the Vandermonde matrix. The Vandermonde matrix for the given reals \(x_1, x_2, \ldots, x_n\) is:
   \[
   V = \begin{pmatrix}
   1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\
   1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\
   \vdots & \vdots & \vdots & \ddots & \vdots \\
   1 & x_n & x_n^2 & \cdots & x_n^{n-1}
   \end{pmatrix}
   \]
   Hadamard's inequality states that for any \(n \times n\) matrix \(A\), the absolute value of the determinant of \(A\) is less than or equal to the product of the Euclidean norms of the columns of \(A\):
   \[
   |\det(A)| \leq \prod_{i=1}^n \|A_i\|
   \]
   where \(A_i\) is the \(i\)-th column of \(A\).

2. **Determinant of the Vandermonde Matrix**:
   The determinant of the Vandermonde matrix \(V\) is given by:
   \[
   \det(V) = \prod_{1 \leq i < j \leq n} (x_i - x_j)
   \]
   Squaring both sides, we get:
   \[
   (\det(V))^2 = \left( \prod_{1 \leq i < j \leq n} (x_i - x_j) \right)^2
   \]

3. **Euclidean Norms of the Columns**:
   The Euclidean norm of the \(i\)-th column of \(V\) is:
   \[
   \|V_i\| = \sqrt{1^2 + x_i^2 + x_i^4 + \cdots + x_i^{2(n-1)}}
   \]
   Summing over all columns, we get:
   \[
   \prod_{i=0}^{n-1} \left( \sum_{j=1}^n x_j^{2i} \right)
   \]

4. **Applying Hadamard's Inequality**:
   Applying Hadamard's inequality to the Vandermonde matrix \(V\), we get:
   \[
   \left( \prod_{1 \leq i < j \leq n} (x_i - x_j) \right)^2 \leq \prod_{i=0}^{n-1} \left( \sum_{j=1}^n x_j^{2i} \right)
   \]

5. **Equality Condition**:
   The equality in Hadamard's inequality holds if and only if the columns of the matrix are orthogonal. For the Vandermonde matrix, this implies that:
   \[
   \sum_{j=1}^n x_j^k = 0 \quad \text{for} \quad k = 1, 2, \ldots, 2n-3
   \]
   For \(n = 2\), this reduces to:
   \[
   x_1 + x_2 = 0
   \]
   For \(n \geq 3\), we need to show that \(x_1 = x_2 = \cdots = x_n = 0\).

6. **Newton's Identities**:
   Using Newton's identities, we can express the power sums in terms of the elementary symmetric polynomials. For \(n > 2\), the conditions:
   \[
   \sum_{\text{sym}} x_1 \cdots x_j = 0 \quad \forall j = 1, \ldots, n
   \]
   imply that the polynomial with roots \(x_1, x_2, \ldots, x_n\) is \(x^n = 0\), which means:
   \[
   x_i = 0 \quad \text{for each} \quad i
   \]

Conclusion:
\[
\boxed{\left( \prod_{1 \leq i < j \leq n} (x_i - x_j) \right)^2 \leq \prod_{i=0}^{n-1} \left( \sum_{j=1}^n x_j^{2i} \right)}
\]
The equality holds if and only if \(x_1 = x_2 = \cdots = x_n = 0\) for \(n \geq 3\), and \(x_1 + x_2 = 0\) for \(n = 2\).