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.
 A: 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 $\sum ab^2 \le \dfrac{4}{27}(\sum a)^3 - abc \le \dfrac{4}{27}(\sum a)^3 = \dfrac{27}{2}$.
A: Point to note: The minimum value is not achieved when $a=b=c=\frac{3}{2}$, which gives a value of $\frac{4}{51}$. The minimum value is achieved at $a=\frac{3}{2}, b=3, c=0$, which gives $\frac{2}{27}$. Note that $\frac{4}{51}>\frac{2}{27}$. (and cyclic permutations)
We shall smooth towards this equality case. 
Note: There was an error with the smoothing approach, and it has been removed. The following only addresses $c=0$.
When $c=0$, we are left to minimise $$\frac{a}{b^3+54}+\frac{b}{54}=\frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54}$$
where $0 \leq b \leq \frac{9}{2}$. This is now a 1 variable inequality, which is easy to work with. Now, 
\begin{align}
& \frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54} \geq \frac{2}{27} \\
\Leftrightarrow &54(\frac{9}{2}-b)+b(b^3+54) \geq 4(b^3+54) \\
\Leftrightarrow &b^4-4b^3+27 \geq 0 \\
\Leftrightarrow &(b-3)^2(b^2+2b+3) \geq 0
\end{align}
The last inequality is clearly true, so the minimum value is $\frac{2}{27}$, when $c=0, b=3, a=\frac{3}{2}$ (and cyclic permutations)
Note: The following shows the motivation to getting the value $\frac{2}{27}$, with equality case $b=3$. Some parts may be just based on intuition and not necessarily rigorous.
Motivation:
Let us suppose that the minimum value is $k$; expanding should give us a quartic, which should have a repeated root at the value of $b$ which gives $k$. (Intuitively) 
\begin{align}
& \frac{\frac{9}{2}-b}{b^3+54}+\frac{b}{54} \geq k \\
\Leftrightarrow &54(\frac{9}{2}-b)+b(b^3+54) \geq 54k(b^3+54) \\
\Leftrightarrow &b^4-54kb^3+243-54^2k \geq 0 \\
\end{align}
Let's do the substitution $l=54k$ to simplify the coefficients, so $$b^4-lb^3+(243-54l) \geq 0$$ We perform another substitution $b=3x$ to simplify further; now $0 \leq x \leq \frac{3}{2}$, and the inequality becomes $$81x^4-27lx^3+(243-54l) \geq 0$$, or equivalently $$3x^4-lx^3+(9-2l) \geq 0$$
We want this to have a repeated root, so we differentiate, and get $12x^3-3lx^2$, which has roots $x=0$ and $x=\frac{l}{4}$. If $x=0$ were to be a repeated root, we would have $l=\frac{9}{2}$ and the quartic inequality becomes $3x^4 \geq \frac{9}{2}x^3$, which fails for $0<x<\frac{9}{2}$. 
Therefore let us try to make $x=\frac{l}{4}$ as the repeated root. We would then require $$0=\left(\frac{l}{4}\right)^3(3(\frac{l}{4})-l)+(9-2l)=9-2l-\frac{l^4}{256}$$
Since I want to avoid calculations as much as possible, let us substitute $m=\frac{l}{4}$. ($m$ is also the value of $x$ where there is a repeated root) Then 
$$m^4+8m-9=0$$
$$(m-1)(m^3+m^2+m+9)$$
We want non-negative $m$, so let's take $m=1$. Then $k=\frac{l}{54}=\frac{4m}{54}=\frac{2}{27}$, and there is a repeated root at $b=3m=3$. Now the quartic polynomial inequality is easy to prove; since we know now that $b=3$ is a repeated root, and thus can factor out $(b-3)^2$ to get a quadratic.
