Given $|u|^2_1 + |v|_1^2 - 2\langle u, v\rangle$, where $|a|_1$ is the one norm defined by $|a|_1 = \sum_i \text{abs}(a_i)$, can one "complete the square" somehow and express it as a single 1-norm quantity (at least an inequality, if not an equality)?
The analogy in case of the two norm is $|u|^2_2 + |v|_2^2 - 2\langle u, v\rangle = |u -v|^2_2$ but I cannot see how to come up with a simpler expression in the case of the 1 norm.
One possibility: $$|u|_1^2+|v|_1^2-2\langle u,v\rangle \le |u|_1^2+|v|_1^2+2|u|_2|v|_2\le |u|_1^2+|v|_1^2+2|u|_1|v|_1=(|u|_1+|v|_1)^2=|w|_1^2,$$ where $w$ is the vector/sequence $(|u_i|+|v_i|)_{i\in\mathbb{N}}$.
Moreover, in $\mathbb{R}^n$, $$|u|_1^2+|v|_1^2-2\langle u,v\rangle\le n|u|_2^2+n|v|_2^2+2n|\langle u,v\rangle| \le n|u\pm v|_2^2\le n|u\pm v|_1^2$$ where $|\langle u,v\rangle|=\pm\langle u,v\rangle$.