15. (25 points) As shown in Figure 1, the parabola $y=a x^{2}$ passes through the point $P(-1,1)$. A line $l$ with a positive slope is drawn through the point $Q\left(-\frac{1}{2}, 0\right)$, intersecting the parabola at points $M$ and $N$ (point $M$ is between $Q$ and $N$). A horizontal line is drawn through $M$, intersecting $O P$ and $O N$ at points $A$ and $B$, respectively. The areas of $\triangle P M A$ and $\triangle O A B$ are denoted as $S_{1}$ and $S_{2}$, respectively. Compare $S_{1}$ with $3 S_{2}$, and provide a reason.
15. Given that the parabola $y=a x^{2}$ passes through the point $P(-1,1)$, we get $a=1$.
Thus, the equation of the parabola is $y=x^{2}$.
Let the line $l: y=k\left(x+\frac{1}{2}\right)(k>0)$.
By combining the equation of line $l$ with the parabola, we get
$$
2 x^{2}-2 k x-k=0 \text {. }
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
x_{1}1-\frac{1}{4}=\frac{3}{4}$.
Note that,
$$
S_{1}=\frac{1}{2}|A M|\left(1-y_{1}\right), S_{2}=\frac{1}{2}|A B| y_{1} \text {. }
$$

Therefore, $S_{1}>3 S_{2}$.