<s>[INST] 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.
 [/INST] By Cauchy-Schwarz,
\begin{align}
 \left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right)(a+b+c) & \geq \left(\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab}\right) ^2 \\
& =  \left (\dfrac{a^2+b^2+c^2}{abc} \right)^2 \\
& = \left (\dfrac{3abc}{abc} \right)^2 \\
& = 9. \\
\end{align}
Thus, dividing both sides of the preceding inequality by $a+b+c$, we have that
$\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right) \geq \dfrac{9}{a+b+c}$, and we are done.
</s>