4.9.

From the problem, transform and rearrange the original inequality to
$$y^{2}-(15-a) x y+a x^{2} \geqslant 0 \text {. }$$

Consider equation (1) as a quadratic inequality in $y$. Since equation (1) holds for all real numbers $y$, we have
$$\Delta=[-(15-a) x]^{2}-4 a x^{2} \leqslant 0 \text {, }$$

which simplifies to $\left(a^{2}-34 a+225\right) x^{2} \leqslant 0$.
Since $x^{2}>0$, it follows that
$$a^{2}-34 a+225 \leqslant 0 \text {. }$$

Solving this, we get $9 \leqslant a \leqslant 25$.
Therefore, the minimum value of $a$ is 9.