I know that for a collection of random variables $\{x_i\}_{i=1}^n$, I have $$\sum_{i,j=1}^nE(x_ix_j)=\sum_{i=1}^{n}E(x_i^2)+2\sum_{i<j}E(x_ix_j). \quad (a)$$ Then triangle inequality gives $$\Big\lvert\sum_{i,j=1}^nE(x_ix_j)\Big\rvert\leq \sum_{i=1}^{n}\lvert E(x_i^2)\rvert+2\sum_{i<j}\lvert E(x_ix_j)\rvert. \quad (b)$$ My question is: does the following hold? $$\sum_{i,j=1}^n\Big\lvert E(x_ix_j)\Big\rvert\leq \sum_{i=1}^{n}\lvert E(x_i^2)\rvert+2\sum_{i<j}\lvert E(x_ix_j)\rvert. \quad (c)$$
My attempt
From the inequality (b), it is not clear if the same upper bound holds for $\sum_{i,j=1}^n\lvert E(x_ix_j)\rvert$. But looking at the first equality (a), I see that the tuple of indices $(i,j)\in[n]^2$ can be equally partitioned as $$\{(i,j)\in[n^2]:i=j\}\cup \{(i,j)\in[n^2]:i>j\}\cup\{(i,j)\in[n^2]:i<j\}.$$ Since $\lvert E(x_ix_j)\rvert=\lvert E(x_jx_i)\rvert$, I would also have $$\sum_{i,j=1}^n \lvert E(x_ix_j)\rvert=\sum_{i=1}^{n}\lvert E(x_i^2)\rvert +2\sum_{i<j}\lvert E(x_ix_j)\rvert,$$ which implies that (c) holds.
Thanks in advance!
The RHS of $$ \sum_{i,j=1}^nE(x_ix_j)=\sum_{i=1}^{n}E(x_i^2)+2\sum_{1\leqslant i<j\leqslant n}E(x_ix_j)\tag1 $$ is just a reordering of the LHS, therefore $$ \sum_{i,j=1}^n|E(x_ix_j)|=\sum_{i=1}^{n}|E(x_i^2)|+2\sum_{1\leqslant i<j\leqslant n}|E(x_ix_j)|\tag2 $$ There is nothing to prove here more than the use of the commutativity of the sum of real numbers, but this was already used in (1). The unique difference between (1) and (2) is that we use in (2) absolute values of it addends.
In general we have that $$ \sum_{i,j=1}^n a_{i,j}=\sum_{i=1}^{n}a_{i,i}+\sum_{1\leqslant i<j\leqslant n}a_{i,j}+\sum_{1\leqslant i<j\leqslant n}a_{j,i}\tag3 $$ for arbitrary complex numbers $a_{i,j}$. In your case we have that $E(x_ix_j)=E(x_jx_i)$ so $$ \sum_{1\leqslant i<j\leqslant n}E(x_ix_j)+\sum_{1\leqslant i<j\leqslant n}E(x_j,x_i)=2\sum_{1\leqslant i<j\leqslant n}E(x_ix_j)\tag4 $$