Difficult Cauchy Problem Let $a, b, c&gt;0$ 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.
$$\sum{a\over bc}=3$$
$$\sum{a\over b^2c^2}\times(a+b+c)\geq(\sum{a\over bc})^2$$ by Cauchy.
[enter image description here][1]
[1]: https://i.stack.imgur.com/VF8FZ.jpg            hi this is my answer is it true?