a+b+c=9/2, and let x=a/(b3+54)+b/(c3+54)+c/(a3+54). Also a,b,c≥0. Note that c depends on a and b and the expression is symmetrical in a, b and c, so the expression is effectively in two variables, not three. c can be replaced by 9/2-a-b.
When c=0 x=a/(b3+54)+b/54 and b=9/2-a. It can be shown that, when a=3/2, b=3, x=(3/2)/(27+54)+3/54=1/54+3/54=2/27 at minimum.
If we increase c to 0.5, a+b=4, and x=a/(b3+54)+b/(⅛+54)+½/(a3+54). It can be shown that, when a=62/45, b=118/45, and x>2/27.
Therefore, it would appear that c=0 is a requirement to minimise x, because c was only increased by a small amount and the value 2/27 was exceeded. Since there is symmetry between a, b and c, this implies that one of the variables needs to be zero, while the other two have values 3/2 and 3 and the minimum value of the expression is 2/27.