Recall that a lattice is a partially ordered set $E$ for which any pair $a,b\in E$ admits a least upper bound and a greatest lower bound. Remark that given four elements $a_i,b_j$ ($j=1,2$), in order that there exists a fifth one $x\in E$ such that $a_i\le x\le b_j$ for all $i,j$, it is necessary and sufficient that $a_i\le b_j$ for every $i,j$. Just take $x=\max\{a_1,a_2\}$ or $\min\{b_1,b_2\}$ or any element in between.
Sadly, or happily, the space $\mathbb H_n$ of Hermitian matrices, equipped with its natural order $\prec$ (Loewner order) is not a lattice. It is not hard to construct $A_1,\ldots,B_2\in\mathbb H_2$ such that $A_i\prec B_j$ for every $i,j=1,2$, but there does not exist an $X\in \mathbb H_2$ such that $A_i\prec X\prec B_j$ for every $i,j$.
Is there a more or less explicit characterization of those quadruplets $A_1,\ldots,B_2\in\mathbb H_n$ such that $A_i\prec B_j$ for every $i,j=1,2$, for which an $X\in\mathbb H_n$ exists such that $A_i\prec X\prec B_j$ for every $i,j$ ?
The following inequality is a necessary condition, though not an accurate one. Let $H\sharp K$ denote the geometrical mean of positive definite matrices. Recall thet it is the greatest hermitian matrix such that $$\begin{pmatrix} H & H\sharp K \\ H\sharp K & K\end{pmatrix}$$ is positive semi-definite. The notion extends by continuity when $H$ and $K$ are only semi-definite. This mean is a monotonous function of $H$ and $K$ under the Loewner order. If $X$ exists as above, then $$(B_j-A_1)\sharp (B_j-A_2)\succ(B_j-X)\sharp (B_j-X)=B_j-X,$$ and likewise $$(B_2-A_i)\sharp (B_1-A_i)\succ (X-A_i)\sharp (X-A_i)=X-A_i.$$ Therefore we must have the NC $$(B_j-A_1)\sharp (B_j-A_2)+(B_2-A_i)\sharp (B_1-A_i)\succ B_j-A_i,$$ for every $i,j=1,2$. That this is not sharp is clear when we think in terms of real numbers (this applies if the matrices differ only form $\mu_{i,j}I_n$): it becomes $$\sqrt{(b_j-a_1)(b_j-a_2)}+\sqrt{(b_2-a_i)(b_1-a_i)}\ge b_j-a_i,$$ where the left-hand side is about twice as big as the right-hand side.
