Is this necessary condition sufficient for "geometric realization"? Let $0<\sigma_1 < \sigma_2$ be fixed positive reals satisfying $\sigma_1 \sigma_2=1$.
Let $0<a \le b$ satisfy $ab \ge 1$. I am looking for necessary and sufficient conditions on $a,b$ that are equivalent to the existence of four positive real numbers
$L_1,L_2,L_3,L_4 \ge 0$ satisfying $\sum L_i=4$, and
$$
\sigma_1 \max(L_1,L_2) \le a\le  \sigma_2 \min(L_1,L_2), \tag{1}
$$
$$
\sigma_1 \max(L_3,L_4) \le b\le  \sigma_2 \min(L_3,L_4). \tag{2}
$$
$a \ge \sigma_1 \max(L_1,L_2) \Rightarrow 2a \ge \sigma_1 (L_1+L_2)$, and similarly $2b \ge \sigma_1 (L_3+L_4)$, so
$2(a+b) \ge 4\sigma_1$. Similarly, $2(a+b) \le 4\sigma_2$.
In other words, we have proved that
$$
\sigma_1\le \frac{a+b}{2} \le\sigma_2, \tag{3}
$$
so this is a necessary condition for the existence of such $L_i$. Is it also sufficient?
If not, what could be a sufficient condition?
Comment: The $\sigma_i$ are part of the data of the problem. They are given. The conditions on $a,b$ should be in terms of the $\sigma_i$.
**If $\sigma_1 = \sigma_2=1$, then conditions $(1),(2)$ indeed reduce to condition $(3)$ $\frac{a+b}{2}=1$.

This  problem came from trying to map a square of edge length $1$ onto a parallelogram.
 A: The condition
$$\sigma_1\max(L_1,L_2)\leq a\leq\sigma_2\min(L_1,L_2)$$
is equivalent with
$$\max(L_1,L_2)\leq{a\over\sigma_1}\qquad\wedge\qquad\min(L_1,L_2)\geq{a\over\sigma_2}\ ,$$
hence with
$${a\over\sigma_2}\leq \  L_1, L_2\ \leq{a\over\sigma_1}\ .$$
It follows that exactly the numbers $$s\in\left[{2a\over\sigma_2}, {2a\over\sigma_1}\right]$$
can be realized as sums  of admissible $L_1$, $L_2$. Similarly, the condition
$$\sigma_1\max(L_3,L_4)\leq b\leq\sigma_2\min(L_3,L_4)$$
implies that exactly the numbers
$$t\in\left[{2b\over\sigma_2}, {2b\over\sigma_1}\right]$$
can be realized as sums of admissible $L_3$, $L_4$. It follows that exactly the numbers
$$z\in\left[{2(a+b)\over\sigma_2}, {2(a+b)\over\sigma_1}\right]$$
can be realized as sums $\sum_{i=1}^4 L_i$ of admissible $L_i$. Since we want
such a sum to be $4$ it is necessary and sufficient that
$${2(a+b)\over\sigma_2}\leq4\leq {2(a+b)\over\sigma_1}\ .$$
This is equivalent with your condition
$$\sigma_1\leq{a+b\over2}\leq\sigma_2\ .$$
