3-Heisenberg-Robertson-Schrodinger Uncertainty Principle

Let $\mathcal{X}$ be a 3-product space. Let $A: \mathcal{D}(A)\subseteq \mathcal{X}\to \mathcal{X}$, $B: \mathcal{D}(B)\subseteq \mathcal{X}\to \mathcal{X}$ and $C: \mathcal{D}(C)\subseteq \mathcal{X}\to \mathcal{X}$ be possibly unbounded 3-self-adjoint operators. Then for all \begin{align*} x \in \mathcal{D}(ABC)\cap\mathcal{D}(ACB) \cap \mathcal{D}(BAC)\cap\mathcal{D}(BCA) \cap \mathcal{D}(CAB)\cap\mathcal{D}(CBA) \end{align*} with $\langle x, x, x \rangle =1$, we show that \begin{align*} (1)\quad \quad \Delta _x(3, A) \Delta _x(3, B) \Delta _x(3, C)\geq |\langle (ABC-a BC-b AC-c AB)x, x, x\rangle +2abc|, \end{align*} where \begin{align*} \Delta _x(3, A):= \|Ax-\langle Ax, x, x \rangle x \|, \quad a:= \langle Ax, x, x \rangle, \quad b := \langle Bx, x, x \rangle, \quad c := \langle Cx, x, x \rangle. \end{align*} We call Inequality (1) as 3-Heisenberg-Robertson-Schrodinger uncertainty principle. Classical Heisenberg-Robertson-Schrodinger uncertainty principle (by Schrodinger in 1930) considers two operators whereas Inequality (1) considers three operators.