Foci of Ellipse lies on Hyperbola and vice-versa Let the foci of the hyperbola $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$  , (A,B > 0) be vertices of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$  , (a, b > 0) and foci of ellipse be vertices of hyperbola. Let eccentricities of the ellipse and hyperbola be $e_1$ & $e_2$ , respectively $L_1$ and $L_2$ are length of Latus rectum of ellipse and hyperbola respectively. Then find the minimum value of $[e_1+e_2]$ (where [.] denotes greatest integer function)
My approach is as follow
Let $ {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} $
Foci of the ellipse are the vertices of the hyperbola and vice-versa then we get the following
$\frac{{{A^2} + {B^2}}}{{{a^2}}} = 1;\frac{{{a^2} - {b^2}}}{{{A^2}}} = 1 \Rightarrow {a^2} - {A^2} = {b^2}$ and ${B^2} = {a^2} - {A^2}$ hence $b=B$
${e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}}  \Rightarrow {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{b^2}}}{{{a^2} - {b^2}}}} $
Not able to proceed further

Further you can write,
$e_1 = \sqrt{\dfrac{a^2-b^2}{a^2}} , e_2 = \sqrt{\dfrac{a^2}{a^2-b^2}}=\dfrac1{e_1} $
Now we have, $e_1+\dfrac1{e_1}$ where $0<e_1<1 $. For the Floor function to achieve minimum value, $e_1$ must be as close as possible to $1$.
For $0<e_1<1$, $~~2<e_1+\dfrac{1}{e_1}<\infty$
So, $\min\left(\Big{\lfloor}e_1+\frac1{e_1}\Big{\rfloor}\right) =2$
