Simple inequality proof problem of my authorship with fixed sum $a+b+c=3$. Prove that for positive numbers $a,b$ and $c$ such that $a+b+c=3$ following inequality holds:
$\sqrt[3]{\frac{1}{3a^2(8b+1)}}+\sqrt[3]{\frac{1}{3b^2(8c+1)}}+\sqrt[3]{\frac{1}{3c^2(8a+1)}}\ge 1$.
There are at least 3 different solutions that I found, so feel free to solve it even if somebody has already posted a solution.

Using AM-GM and Cauchy-Schwarz:
$$
\begin{aligned}
\frac{1}{3}\cdot LHS &= \sum \frac{1}{\sqrt[3]{9a\cdot 9a \cdot (8b+1)}} \\
&\geq \sum \frac{3}{9a+9a+8b+1} \\
&= 3\sum \frac{1}{18a+8b+1} \\
&\geq 3\cdot \frac{9}{26(a+b+c)+3} \\
&= \frac{1}{3}
\end{aligned}
$$
Equality occurs when $a=b=c=1$.
