Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.
I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.
I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

Proposed inequality can be rewriten like this:$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$
By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2$$ and we are done!
