Let $A,B,C,D$ be bounded operators on a Hilbert space $\mathcal{H}$. I know that $$ \|AB\| \leq \|A\otimes B\| $$ where $\otimes$ is the tensor product and $\|\cdot\|$ is the operator norm. I wonder if this extends to linear combinations. That is, is the following generalization true: $$ \|AB + CD\| \leq \|A\otimes B + C\otimes D\| ? $$
My intuition is that it is ``harder'' for $A\otimes B$ to cancel $C\otimes D$ than it is for $AB$ to cancel $CD$, so I think it might be true.
Let's assume that the Hilbert space has at least dimension $2$ (in dimension $1$ the inequality is obviously true) and pick three orthonormal vectors $e_1, e_2$. Then consider the rank $1$ projections $$ A= \vert e_1 \rangle \langle e_1 \vert, B= \vert e_1 \rangle \langle e_1 \vert, C= \vert e_1 \rangle \langle e_2 \vert, D = \vert e_2 \rangle \langle e_1 \vert. $$ Then we get $$ AB+CD=2 \vert e_1 \rangle \langle e_1 \vert. $$ Clearly this has operator norm $2$. Extend our family to a Hilbert basis $(e_j)_{j\in \Lambda}$ of $H$. Then we compute $$ (A\otimes B+C\otimes D)(\sum_{i,j\in \Lambda} c_{i,j} \, e_i \otimes e_j) = c_{1,1} \, e_1 \otimes e_1+ c_{2,1} \, e_1 \otimes e_2.$$ By orthogonality we get $$ \Vert c_{1,1} \, e_1 \otimes e_1+ c_{2,1} \, e_1 \otimes e_2 \Vert = \sqrt{\vert c_{1,1}\vert^2+\vert c_{2,1}\vert^2} \leq \left( \sum_{i,j\in \Lambda} \vert c_{i,j}\vert^2\right)^{1/2} = \Vert \sum_{i,j\in \Lambda} c_{i,j} \, e_i \otimes e_j \Vert. $$ Thus, $ \Vert A\otimes B + C \otimes D \Vert =1$ and hence $$ \Vert AB+CD \Vert =2 > 1 = \Vert A\otimes B + C \otimes D \Vert. $$
As requested by @SeverinSchraven, here is an elaboration of my comments. What I had in mind was something like the following:
Consider two $C^\ast$-algebras $\mathcal{A}$, $\mathcal{B}$. In $C^\ast$-algebra theory, there are, in general, multiple norms that can be equipped on the tensor product $\mathcal{A} \otimes \mathcal{B}$ to make it, after completion, into a $C^\ast$-algebra. Two canonical choices are the smallest, minimal (also called spatial) tensor norm $\otimes_\min$; and the largest, maximal tensor norm $\otimes_\max$. A $C^\ast$-algebra $\mathcal{A}$ is called nuclear if for all $C^\ast$-algebra $\mathcal{C}$, the minimal and maximal tensor norms on $\mathcal{A} \otimes \mathcal{C}$ coincide. It is a classic result that there are non-nuclear $C^\ast$-algebras, for example the reduced group $C^\ast$-algebra $C^\ast_r(G)$ (or full group $C^\ast$-algebra $C^\ast(G)$) of a discrete, non-amenable group $G$ (for example any non-abelian free group). Also, $\mathbb{B}(l^2)$ is non-nuclear. Regardless, the point is there exists some $C^\ast$-algebras $\mathcal{A}$ and $\mathcal{B}$ s.t. $\mathcal{A} \otimes_\min \mathcal{B} \neq \mathcal{A} \otimes_\max \mathcal{B}$. However, basically by definition, there always exists a contractive $\ast$-homomorphism $\mathcal{A} \otimes_\max \mathcal{B} \to \mathcal{A} \otimes_\min \mathcal{B}$ which is the identity map on the algebraic tensor product $\mathcal{A} \otimes \mathcal{B}$. $\mathcal{A} \otimes_\min \mathcal{B} \neq \mathcal{A} \otimes_\max \mathcal{B}$ then means there are some $A_1, \cdots, A_n \in \mathcal{A}$ and $B_1, \cdots, B_n \in \mathcal{B}$ s.t. $\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\min \mathcal{B}} < \|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\max \mathcal{B}}$.
Let $H$ be a Hilbert space on which $\mathcal{A} \otimes_\max \mathcal{B}$ acts faithfully. Both $\mathcal{A}$ and $\mathcal{B}$ canonically (and isometrically) embeds into $\mathcal{A} \otimes_\max \mathcal{B}$, so we can regard $A_i$, $B_i$ as operators on $H$. Then,
$$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\max \mathcal{B}} = \|\sum_{i=1}^n A_iB_i\|_{\mathbb{B}(H)}$$
On the other hand, again basically by definition of the minimal tensor norm,
$$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\min \mathcal{B}} = \|\sum_{i=1}^n A_i \otimes B_i\|_{\mathbb{B}(H \otimes H)}$$
So you have $\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathbb{B}(H \otimes H)} < \|\sum_{i=1}^n A_iB_i\|_{\mathbb{B}(H)}$.
