Definitions: We generate random triangles using various distribution (uniform, gaussian etc) for $\theta$ to get the polar coordinates of the vertices $\cos \theta,\sin \theta$ in a circle of radius $r = 1$ and define:
Ordered mean-triangle of equal area: Let $a_x, b_x$ and $c_x$ be mean of the smallest, the intermediate and the longest side of all triangles of area $x$. The triangle with sides $a_x, b_x$ and $c_x$ is called the ordered mean-trinagle of equal area $x$.
Random mean-triangle of equal area: Let $p_x, q_x$ and $r_x$ be mean of the side taken in no particular order of all triangles of area $x$. The triangle with sides $a_x, b_x$ and $c_x$ is the random mean-trinagle of equal area $x$.
Let $\Delta_{a,b,c}$ be the area of a triangle of sides $a,b$ and $c$. Experimental data shows that, the area of both the ordered mean-triangle and the random mean-triangle is greater than that of $x$. More specifically, we have $$ x \le \Delta_{a_x,b_x,c_x} \le \Delta_{p_x,q_x,r_x} \tag 1 $$
with equality occuring only if $x = 0$ or $\displaystyle x = \frac{3\sqrt{3}}{4}$. The graph below plots the area $x$ in the $X$-axis and the differences $\Delta_{a_x,b_x,c_x} - x$ and $\Delta_{p_x,q_x,r_x} - x$ on the $Y$-axis.
Question: Is the area of a triangle is less than that of its mean triangle of equal area?