By the symmetry of equation (1), we first make a common substitution. Let $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$, then equation (1) can be rewritten as
$$x f\left(x^{2}+8 y z\right)+y f\left(y^{2}+8 z x\right)+z f\left(z^{2}+8 x y\right) \geqslant 1$$

where $f(t)=\frac{1}{\sqrt{t}}$. By the convexity of $f$ and $x+y+z=1$, we have
$$\begin{aligned} & x f\left(x^{2}+8 y z\right)+y f\left(y^{2}+8 z x\right)+z f\left(z^{2}+8 x y\right) \\ \geqslant & f\left(x\left(x^{2}+8 y z\right)+y\left(y^{2}+8 z x\right)+z\left(z^{2}+8 x y\right)\right) \end{aligned}$$

Since $f(t)=\frac{1}{\sqrt{t}}$ is monotonically decreasing and $f(1)=1$, it suffices to prove
$$1 \geqslant x\left(x^{2}+8 y z\right)+y\left(y^{2}+8 z x\right)+z\left(z^{2}+8 x y\right)$$

In fact,
$$\begin{aligned} & (x+y+z)^{3}-x\left(x^{2}+8 y z\right)-y\left(y^{2}+8 z x\right)-z\left(z^{2}+8 x y\right) \\ = & 3\left[x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}\right] \geqslant 0 . \end{aligned}$$

Therefore, equation (1) holds.