15. Let $m>0$, if for any set of positive numbers $a, b, c$ satisfying $a b c \leqslant \frac{1}{4}$ and $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}<m$, there always exists a triangle with side lengths $a, b, c$, find the maximum value of the real number $m$, and explain the reason.
When $m>9$, take $a=b=\frac{1}{2}, c=1$, then $a, b, c$ cannot form a triangle $\Rightarrow m \leqslant 9$. On the other hand, suppose $m \leqslant 9$, there exist $a \leqslant b \leqslant c$ that cannot form a triangle, thus $a+b \leqslant c$. Let $d=\frac{1}{4 a b} \geqslant c$, and $d^{2} \geqslant c^{2} \geqslant(a+b)^{2} \geqslant 4 a b=\frac{1}{d} \Rightarrow d \geqslant 1$.
Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{d^{2}}=\left(\frac{1}{a}-\frac{1}{b}\right)^{2}+8 d+\frac{1}{d^{2}} \geqslant 8 d+\frac{1}{d^{2}}$ $=9+\frac{8 d^{3}-9 d^{2}+1}{d^{2}}=9+\frac{(d-1)\left(8 d^{2}-d-1\right)}{d^{2}} \geqslant 9$, contradiction.

Therefore, for any set of positive numbers $a, b, c$ that satisfy $a b c \leqslant \frac{1}{4}$ and $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}<9$, there exists a triangle with side lengths $a, b, c$, i.e., the maximum value of the real number $m$ is 9.