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

In the given notation the focus of ellipse is $S_1(ae_1,0)$ and vertex is $V_1(a,0)$.
For the hyperbola $S_2=(Ae_2,0)$ and $V_2=(A,0)$. Given that $S_1=V_2$ and S_2=V_1$.
we get $e_1=A/a$ and $e_2=a/A$ \implies $e_1 e_2=1$, so by AM-GM, we get
$$\frac{e_1+e_2}{2}\ge \sqrt{e_1 e_2} \implies [e_1+e_2]\ge 2.$$
