Let $AA_1$, $BB_1$, $CC_1$ meet the sides of $\triangle A_1B_1C_1$ at $A_2$, $B_2$, $C_2$ respectively.
Let $\frac{AC_1}{C_1B} = \frac{x}{y}$, $\frac{BA_1}{A_1C} = \frac{y}{z}$ ($x, y, z > 0$), then by Ceva's Theorem $\frac{CB_1}{B_1A} = \frac{z}{x}$. Thus the center of mass of the system of mass points
$A(yz)$, $B(zx)$, $C(xy)$
is at $O$. (Since $\frac{yz}{zx} = \frac{C_1B}{AC_1}$, the center of mass of $A(yz)$ and $B(zx)$ is $C_1$, so the center of mass of the whole system lies on $CC_2$; by the same reasoning it also lies on $BB_1$, so it must be their intersection point $O$.)
On the other hand, since the center of mass of $A(\frac{yz}{2})$ and $B(\frac{zx}{2})$ is also $C_1(A_1, B_1)$, similarly $O$ is also the center of mass of the system of mass points
$A_1(\frac{zx + xy}{2})$, $B_1(\frac{xy + yz}{2})$, $C_1(\frac{yz + zx}{2})$.
Suppose for contradiction that $O$ does not lie inside the medial triangle of $A_1B_1C_1$. Then it lies in one of the other three small triangles near $A_1$, $B_1$, $C_1$; without loss of generality assume it is the one near $A_1$, so that $OA_2 \ge OA_1$. But since $O$ is the center of mass of the system of mass points
$A_1(\frac{zx + xy}{2})$, $B_1(\frac{xy + yz}{2})$, $C_1(\frac{yz + zx}{2})$,
the center of mass of $B_1(\frac{xy+yz}{2})$ and $C_1(\frac{yz+zx}{2})$ must lie on the line $OA_1$, and can only be
$A_2$. And $O$ is the center of mass of $A_1(\frac{zx+xy}{2})$ and $A_2(\frac{xy+yz}{2} + \frac{yz+zx}{2})$. Since
$$\frac{xy + yz}{2} + \frac{yz + zx}{2} > \frac{zx + xy}{2}$$
we have $OA_2 < OA_1$, a contradiction!