15. Proof one: Let $S=a+b-c, f(x)=\frac{1}{1-x}, x \in(0, \mathrm{~F})$. The original inequality is equivalent to
$$\frac{a}{S} f(a)+\frac{b}{S} f(b)+\frac{c}{S} f(c) \geqslant \frac{3 \sqrt{3}+3}{2 S}$$

By Jensen's inequality, we have
$$\frac{a}{S} f(a)+\frac{b}{S} f(b)+\frac{c}{S} f(c) \geqslant f\left(\frac{a^{2}+b^{2}+c^{2}}{S}\right)=f\left(\frac{1}{S}\right)$$

By Cauchy-Schwarz inequality, we get $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2}$, hence $S \leqslant \sqrt{3}$, so
$$f\left(\frac{1}{S}\right) \geqslant f\left(\frac{\sqrt{3}}{3}\right)=\frac{3 \sqrt{3}+3}{2}$$