Simple inequality proof problem of my authorship with fixed sum $a+b+c=3$.

Question

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.

Answer 1

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 $a=b=c=1$.

Answer 2

After using AM-GM it's enough to prove that: $$729\geq\prod_{cyc}a^2(8a+1),$$ which is true by AM-GM: $$abc\leq\left(\frac{a+b+c}{3}\right)^3=1$$ and $$(8a+1)(8b+1)(8c+1)\leq\left(\frac{8a+1+8b+1+8c+1}{3}\right)^3=729.$$