Double sum of mixed terms I have a double sum given by
$$ \sum_{i=1}^n \sum_{j=1,\ j\neq i}^n x_i\cdot x_j = 2\sum_{i=1}^n\sum_{j<i}x_i\cdot x_j, $$
where the equality supposedly stems from the fact that
$$ \sum_{i=1}^n\sum_{j<i} x_i\cdot x_j=\sum_{i=1}^n\sum_{j>i} x_i\cdot x_j. $$
I see how the first equation follows from the second, and I see why the second equation holds intuitively (e.g. by writing it down for $n=3$...).
But I cannot think of a proper proof for the second equation. Is there one or is this too "trivial" to be proven?
 A: Given $x_1,x_2,\cdots,x_n$ let $A=(a_{ij})$ denote the square matrix defined by $a_{ij}=x_i\cdot x_j$.
Then $A$ is a symmetric matrix. So the sum of the terms below the main diagonal equals the sum of the terms above the main diagonal.
That is
$$ \sum_{i=1}^n\sum_{j<i} x_i\cdot x_j=\sum_{i=1}^n\sum_{j>i} x_i\cdot x_j. $$
A: Use mathematical induction.
The inductive step goes like this :
$
\begin{array}{rcl}
\sum\limits_{i=1}^{n+1} \sum\limits_{j<i} x_i x_j &=& \sum\limits_{i=1}^{n} \sum\limits_{j<i} x_i x_j + x_{n+1} \sum\limits_{j<n+1} x_j \\
& = & \sum\limits_{i=1}^{n} \sum\limits_{n \geq j>i} x_i x_j + x_{n+1} \sum\limits_{j<n+1} x_j \\
& = & \sum\limits_{i=1}^{n+1} \sum\limits_{j>i} x_i x_j 
\end{array}
$
A: The following representation might be helpful.

We obtain
  \begin{align*}
\color{blue}{\sum_{i=1}^n\sum_{{j=1}\atop{j\ne i}}^nx_ix_j}
&=\sum_{1\leq i<j\leq n}x_ix_j+\sum_{1\leq j<i\leq n}x_ix_j\\
&=\sum_{1\leq j<i\leq n}x_jx_i+\sum_{1\leq j<i\leq n}x_ix_j\\
&=2\sum_{1\leq j<i\leq n}x_ix_j\\
&\color{blue}{=2\sum_{i=2}^n\sum_{j=1}^{i-1}x_ix_j}\\
\end{align*}

Hint: In the expression $\sum_{i=1}^n\sum_{j<i}x_ix_j$ is the inner sum with entry  $i=1$ an empty sum equal to zero.
