Deje $f_j(w)= \sqrt{1-w}\exp\left(w\dfrac{x_j^2}2\right),$ $f'_j(w)=\left(-\dfrac1{2\sqrt{1-w}}+\sqrt{1-w}\dfrac{x_j^2}2\right)\exp\left(w\dfrac{x_j^2}2\right).$
Es máxima cuando $f'_j(w)=0$ ie $x_j^2 = \dfrac1{1-w},$
$$f_j(w) \leq f(w),\quad \text{ where}\quad f(w) = \sqrt{1-w}\exp\dfrac w{2(1-w)}.$$
Derivación de $f(w)$:$\quad f'(w)=\left(-\dfrac1{2\sqrt{1-w}} +\dfrac{\sqrt{1-w}}{2(1-w)^2}\right)\exp\dfrac w{2(1-w)},$
$$f'(w)=\dfrac w{2(1-w)^2}f(w).$$
$f(w)$ mínimo de al $$w = 0,\quad f_{min}(w) = f(0) = 1,$$ y singularidad cuando
$$w\to1,\quad \lim\limits_{w\to 1}f(w) = \infty.$$
Por otro lado, la función de $$g_i(w)=\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right)$$ has maximum at $w\in[0,1]$ when $w=0$, ie $$\max\limits_{w\in[0,1]}\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right) = 1.$$
Así
$$A_{i} \leq \int_0^1 \int_0^1 \left(p-(1-p)\min_wf_i(w)\left[1-\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right)\right]\right) \cdot \prod_{j\neq i}^m (p+(1-p)f(w))dwdp =$$$$\int_0^1 \int_0^1 p(p+(1-p)f(w))^{m-1}dwdp \leq \int_0^1 \int_0^1 pf^{m-1}(w))dwdp = \dfrac12\int_0^1 f^{m-1}(w)dw.$$
$$\boxed{A_{i} \leq \dfrac12\int_0^1 f^{m-1}(w)dw}$$
Sobre el tema principal.
$$A_i-A_j = \int_0^1\int_0^1(((p-(1-p)f_i(w)(1-g_i(w))((p+(1-p)f_j(w))- $$$$((p-(1-p)f_j(w)(1-g_j(w))((p+(1-p)f_i(w)))\cdot\prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp = $$$$ \int_0^1\int_0^1(1-p)(p(f_i(w)g_i(w))-f_j(w)g_j(w)) + (1-p)f_i(w)f_j(w)(g_i(w)-g_j(w)))\cdot \prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp = $$$$ \int_0^1\int_0^1(1-p)(p(h_i(w)-h_j(w)) + (1-p)f_i(w)f_j(w)(g_i(w)-g_j(w)))\cdot \prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp, $$
donde $h_i(w) = f_i(w)\cdot g_i(w).$
Teniendo en cuenta esta fórmula con las funciones
$$f(w,t) = \sqrt{1-w}\exp wt,\quad g(w,t) = \dfrac1{\sqrt{1+w}}\exp\left(-\dfrac{w^2t}{w+1}\right),\quad h(w,t) = f(w,t)g(w,t) = \sqrt{\dfrac{1-w}{1+w}}\exp\dfrac{wt}{w+1}$$ when $\dfrac{x_j^2}2\a \dfrac{x_i^2}2 = t,$ podemos diferencias de cambio en virtud de la integral de acuerdo a los cálculos:
$$\dfrac{h_i(w)-h_j(w)}{t_i-t_j}\to\dfrac{dh}{dt} = \dfrac w{w+1}h(w,t),$$
$$f_i(w)f_j(w)\dfrac{g_i(w)-g_j(w)}{t_i-t_j}\to f^2(w,t)\dfrac{dg}{dt} = -\dfrac{w^2}{w+1}f(w,t)^2g(w,t) = -\dfrac{w^2}{w+1}f(w,t)h(w,t) = -wf(w,t)\dfrac{dh}{dt}.$$
Así
$$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))\dfrac{dh}{dt}\,\prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))\,dwdp.$$
Al $m=2$:
$$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))\dfrac{dh}{dt}\,dwdp = {\dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1((p-p^2)-(1-p)^2w \sqrt{1-w}\exp(wt))\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dwdp}= {\dfrac{x_i^2-x_j^2}{12}\int\limits_0^1(1-2w \sqrt{1-w}\exp(wt))\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dw}.$$
El uso de Wolfram Alpha programa, podemos ver que esta expresión es mayor que $0$ al $t\leq 0.83$ y menos de $0$ al $t\geq 0.84$.
Al $m=3$:
$$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))(p+(1-p)f(w,t_1))\dfrac{dh}{dt}\,dwdp = \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^ 1(p^2(1-p) - p(1-p)^2wf(w,t) + p(1-p)^2f(w,t))+(1-p)^3wf(w,t)f(w,t_1))\dfrac{dh}{dt}\,dwdp = $$$$\dfrac{x_i^2-x_j^2}2\int\limits_0^ 1\left(\dfrac1{12} - \dfrac7{12}wf(w,t) + \dfrac7{12}f(w,t_1)-\dfrac14wf(w,t)f(w,t_1)\right)\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^ 1(6 - 21wf(w,t) + 21f(w,t_1)-9wf(w,t)f(w,t_1))\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^1((7 - 3wf(w,t))(7 + 3f(w,t_1))-43)\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^1((7 - 3w\sqrt{1-w}\exp(wt))(7 + 3\sqrt{1-w}\exp(wt_1))-43)\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dw.$$
El uso de Wolfram Alpha programa en $t=4$, podemos ver que la expresión es mayor que $0$ al $t_1\geq 2.8$ y menos de $0$ al $t_1\leq 2.7$.
Por lo que la hipótesis acerca de la $A_i<A_j$ es malo.
