The left-hand side is indeed always greater than or equal to the right-hand side, i.e. the inequality holds with $\Box = \ge$, as you suspected. With the substitution $f = 1/A_1$, $g= 1/A_2$ the statement becomes:
Let $f, g: [a, b] \to (0, \infty)$ be (Riemann or Lebesgue) integrable. Then
$$
\left( \int_a^b f(x) \, dx \right) \cdot \left( \int_a^b g(x) \, dx \right)
\ge \left( \int_a^b \frac{f(x) g(x)}{f(x)+g(x)} \, dx \right) \cdot \left( \int_a^b (f(x)+g(x)) \, dx \right) \, .
$$
As shown in Integral Ineq. 2 on AoPS, this is a consequence of Jensen's integral inequality for the concave function $B(u, v) = \frac{uv}{u+v}$. For those (like me) who are more familiar with Jensen's inequality for convex or concave functions of a single variable, here is a short proof:
Set
$$
F = \frac{1}{b-a}\int_a^b f(x) \, dx \, , \,
G = \frac{1}{b-a}\int_a^b g(x) \, dx \, .
$$
If $F=0$ then $f(x) = 0$ a.e. on $[a, b]$ and both sides of the inequality are zero, similar if $G= 0$, so we can assume that $F > 0$ and $G > 0$.
$B(u, v) = \frac{uv}{u+v}$ is concave on $(0, \infty) \times (0, \infty)$ (see for example here), so that
$$
B(u, v) \le B(F, G) + \alpha (u-F) + \beta (v-G)
$$
for all $u, v > 0$, where $\alpha = B_u(F, G)$ and $\beta = B_v(F, G)$ are the partial derivatives of $B$ at the point $(F, G)$. For $u =f(x)$, $v=g(x)$ one gets
$$
\frac{f(x)g(x)}{f(x)+g(x)} \le \frac{FG}{F+G} + \alpha(f(x)-F) + \beta(g(x) - G) \, ,
$$
and integrating this over $[a, b]$ gives
$$
\int_a^b \frac{f(x)g(x)}{f(x)+g(x)} \le (b-a) \frac{FG}{F+G}
= \frac{\left( \int_a^b f(x) \, dx \right) \cdot \left( \int_a^b g(x) \, dx \right)}{\int_a^b (f(x)+g(x)) \, dx} \, .
$$
