Show that for real numbers $x_1, x_2, ... , x_n$ we have:
\[ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \dfrac{x_ix_j}{i+j} \geq 0  \] 
When do we have equality?
1. Define the function \( F(x) \) as follows:
   \[
   F(x) = \left( \sum_{i=1}^{n} x_i x^{i-\frac{1}{2}} \right)^2
   \]
   for \( 0 < x \leq 1 \). Notice that \( F(x) = g(x)^2 \) where \( g(x) = \sum_{i=1}^{n} x_i x^{i-\frac{1}{2}} \).

2. Since \( F(x) = g(x)^2 \), it is clear that \( F(x) \geq 0 \) for all \( x \) in the interval \( (0, 1] \) because the square of any real number is non-negative.

3. Integrate \( F(x) \) from 0 to 1:
   \[
   \int_0^1 F(x) \, dx = \int_0^1 \left( \sum_{i=1}^{n} x_i x^{i-\frac{1}{2}} \right)^2 \, dx
   \]

4. Expand the square inside the integral:
   \[
   \int_0^1 \left( \sum_{i=1}^{n} x_i x^{i-\frac{1}{2}} \right)^2 \, dx = \int_0^1 \left( \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j x^{i+j-1} \right) \, dx
   \]

5. Interchange the sum and the integral:
   \[
   \int_0^1 \left( \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j x^{i+j-1} \right) \, dx = \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j \int_0^1 x^{i+j-1} \, dx
   \]

6. Evaluate the integral:
   \[
   \int_0^1 x^{i+j-1} \, dx = \left[ \frac{x^{i+j}}{i+j} \right]_0^1 = \frac{1}{i+j}
   \]

7. Substitute the result of the integral back into the sum:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j \frac{1}{i+j}
   \]

8. Therefore, we have:
   \[
   \int_0^1 F(x) \, dx = \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{x_i x_j}{i+j}
   \]

9. Since \( F(x) = g(x)^2 \geq 0 \), it follows that:
   \[
   \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{x_i x_j}{i+j} \geq 0
   \]

10. Equality holds if and only if \( g(x) \) is identically zero, which means:
    \[
    \sum_{i=1}^{n} x_i x^{i-\frac{1}{2}} = 0 \quad \text{for all} \quad 0 < x \leq 1
    \]
    This implies that all \( x_i \) must be zero.

The final answer is \( \boxed{ \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{x_i x_j}{i+j} \geq 0 } \) with equality if and only if all \( x_i = 0 \).