Proof: Although this problem has been solved using Hölder's inequality, the proof given by Jensen's inequality is still very elegant. Without loss of generality, we assume $a+b+c=1$. Since $f(x)=\frac{1}{\sqrt{x}}$ is a convex function, by Jensen's inequality we get:
$$a f\left(a^{2}+8 b c\right)+b f\left(b^{2}+8 c a\right)+c f\left(c^{2}+8 a b\right) \geq f(M)$$

Here $M=\sum_{c y c} a\left(a^{2}+8 b c\right)=24 a b c+\sum_{c y c} a^{3}$. Therefore, we only need to prove $f(M) \geq 1$ or $M \leq 1$ or
$$24 a b c+\sum_{c y c} a^{3} \leq\left(\sum_{c y c} a\right)^{3} \Leftrightarrow \sum_{c y c} c(a-b)^{2} \geq 0$$

The last inequality is obvious. Equality holds when $a=b=c$.