Formally showing $2\sum_{j=1}^{T-1}\sum_{t=1}^{T-j}x_t x_{t+j} = \sum_{i=1}^{T}x_i( \sum_{j\not=i} x_j)$ Or, if it isn't true please tell me, as I messed up earlier. But I think I may have gotten it correct this time.
Here is what I did; any suggestions (particularly on how I can state things more formally) is much appreciated.
Notice that, for a given $j$, $x_i \in
          \sum_{t=1}^{T-j}x_t x_{t+j}$
                        if $i+j\leq T$. Therefore, each $x_i$ appears in $(T-i)$ terms.
                        Thus, we can write
            $$
    \sum_{j=1}^{T-1}
           \sum_{t=1}^{T-j}x_t x_{t+j}
    = \sum_{i=1}^{T-1} x_i (\sum_{j=i+1}^{T}x_j)
   $$
    But we also have that (using the fact that if $x_j$ must show up in $(j-1)$ terms of the form $x_jx_{i<j}$ for all $i\in \{1,\dots,j-1\}$)
        $$
   \sum_{i=1}^{T-1} x_i (\sum_{j=i+1}^{T}x_j) =
    x_T\sum_{i=1}^{T-1} x_i +x_{T-1}\sum_{i=1}^{T-2} + \dots +x_{3}\sum_{i=1}^2x_i + x_1x_2 =  \sum_{i=1}^{T}x_i ( \sum_{j=1}^{i-i}x_j)
  $$
    Putting this together we have that
        $$
  2\sum_{j=1}^{T-1}\sum_{t=1}^{T-j}x_t x_{t+j} =  
    \sum_{i=1}^{T-1}x_i(\sum_{j=i+1}^T x_j) +
     \sum_{i=1}^T x_i( \sum_{j=1}^{i-1}x_j)
   =  \sum_{i=1}^{T-1}x_i( \sum_{j\not=i} x_j) + x_T\sum_{i=1}^{T-1}x_j
  $$
which can be further simplified to
$$
\sum_{i=1}^{T}x_i( \sum_{j\not=i} x_j)
$$

OPs approach is correct. Here is a somewhat more formal derivation.

We obtain
  \begin{align*}
\sum_{j=1}^{T-1}\sum_{t=1}^{T-j}x_tx_{t+j}&=\sum_{j=1}^{T-1}\sum_{t=j+1}^Tx_{t-j}x_t\tag{1}\\
&=\sum_{1\leq j<t\leq T}x_tx_{t-j}\tag{2}\\
&=\sum_{1\leq j<t\leq T}x_tx_j\tag{3}\\
&=\frac{1}{2}\sum_{{1\leq t,j\leq T}\atop{t\ne j}}x_tx_j\tag{4}\\
&=\frac{1}{2}\sum_{t=1}^{T}\sum_{{j=1}\atop{j\ne t}}^{T}x_tx_j\tag{5}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we shift the index of the inner sum by $j$.

*In (2) we use another representation of (1) to better see the range of summation. We do not change anything.

*In (3) we replace $j \rightarrow t-j$.

*In (4) we use the symmetry when exchanging $i$ with $j$. More explicit
\begin{align*}
\sum_{1\leq j<t\leq T}x_tx_j=\sum_{1\leq t<j\leq T}x_tx_j=\frac{1}{2}\sum_{{1\leq t,j\leq T}\atop{t\ne j}}x_tx_j
\end{align*}

*In (5) we do not change anything. We just use another representation of (4).
