<s>[INST] If $ a+b+c = \frac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$ 
If $a+b+c = \dfrac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $$\dfrac{a}{b^3+54}+\dfrac{b}{c^3+54}+\dfrac{c}{a^3+54} \qquad ?$$

My try: 
$$\begin{align*}
\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54} &= \frac{a^2}{ab^3+54a}+\frac{b^2}{bc^3+54b}+\frac{c^2}{ca^3+54c}\\
&\geq \frac{(a+b+c)^2}{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}
\end{align*}$$
Now, how can I proceed after that? Thanks.
 [/INST] Let $\sum$ denote the cyclic sum in the answer below.  Also, let $f$ denote the expression to be minimised.  Then,
$$f = \sum \frac{a}{b^3+54} = \frac{1}{54}\left(\sum a - \frac{ab^3}{b^3+54}\right)$$
By AM-GM, $\qquad b^3+54 = b^3 + 27 + 27 \ge 27 b$
So $$54f \ge \frac{9}{2} -  \frac{1}{27} \sum ab^2$$
We prove below that $\sum ab^2 \le \frac{27}{2}$.  However this gives
$$f \ge \frac{2}{27}$$
Equality is obtained for cyclic permutations of $(\frac{3}{2}, 3, 0)$, so this is the minimum (as surmised by Ivan Loh).

We prove here that $\sum ab^2  \le \dfrac{4}{27}(\sum a)^3 = \dfrac{27}{2}$. 
Let $x, y, z$ be a permutation of $a, b, c$; s.t. $x \ge y \ge z \ge 0$.
Then $\sum x = \sum a$ and $xy \ge zx \ge yz$.
Then by rearrangement inequality, we have  
$\sum ab^2 \le x(xy) + y(zx) + z(yz) = y(x+z)^2 - xyz$  
But $y(x+z)^2 = \dfrac{1}{2} 2y (x+z)^2 \le \dfrac{1}{2} \dfrac{(2y + x + z + x + z)^3}{27} = \dfrac{4}{27}(\sum x)^3$  
So $\sum ab^2 \le \dfrac{4}{27}(\sum a)^3 - abc \le \dfrac{4}{27}(\sum a)^3 = \dfrac{27}{2}$.
</s>