Non-trivial bin-packing instance with 5 objects
Let's look at two algorithms: Best Fit Ordered, and First Fit Ordered. Both take the items in descending order and put them into a non-empty bin if possible, or into a new empty bin if they don't fit into any existing bin. Best Fit Ordered puts an item into the filled bin with least available space, First Fit Ordered puts an item into the first filled bin with enough space. 

Question: Are there five items A ≥ B ≥ C ≥ D ≥ E such that BFO or FFO are not optimal? 

If all items fit into one bin then BFO and FFO will fit them into one bin, because both will just put items in the first bin. Both are optimal.

If five bins are optimal then BFO and FFO cannot be non-optimal, since they can't use more than five bins. 

If four bins are optimal then there are two items fitting into one bin, therefore D and E fit into one bin, therefore when BFO and FFO finally try to find a bin for E, either four bins are filled and E fits into the bin with D, or there are only three bins filled and E goes into the fourth bin. 

So the only cases to consider are those where two or three bins are optimal. I'll consider cases systematically:

A, B and another fit -> 2 bins, BFO and FFO are optimal
A, B fit, C, D, E fit -> 2 bins, BFO and FFO are optimal
A, B fit, C, D, E don’t fit -> 3 bins required, BFO and FFO are optimal. 

We now assume A and B don't fit into a bin, so A goes in the first and B in the second:

A, C and another fit -> 2 bins, BFO and FFO are optimal
A, C fit, B, D, E fit -> 2 bins, BFO and FFO are optimal
A, C fit, B, D, E don’t fit: A, D, E don’t fit, A, D or A, E doesn’t help -> 3 bins required -> BFO and FFO are optimal.

We now assume neither A and B, nor A and C, fit into one bin. If B and C don't fit, then A, B, C go in three different bins. 3 bins are optimal, so D and E can be made to fit, and BFO and FFO will both find a way. 

So we assume B and C fit in one bin. Both BFO and FFO will fit them in one bin. D  E must also fit into one bin, so if three bins are optimal, then BFO and FFO will find a solution. 

If two bins are optimal, then one bin must contain A, one must contain B and C. D and E fit, so BFO and FFO will fit them.