05AB1E, 15 14 13 bytes

εDO5÷FR]{`-Pd

-1 byte thanks to @Neil.

Try it online or verify all test cases.

Explanation:

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 {       # Sort the list of pairs from lowest to highest
  `      # Pop and push both pairs separated to the stack
   -     # Subtract the values of the pairs from one another at the same indices
    P    # Take the product of those two values
     d   # And check that it's non-negative / >=0 (thus no score is decreasing)
         # (after which the result is output implicitly)

05AB1E, 15 14 12 bytes

εDO5÷FR]`-Pd

-2 bytes thanks to @Neil.

Try it online or verify all test cases.

Explanation:

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 `       # Pop and push both pairs separated to the stack
  -      # Subtract the values of the pairs from one another at the same indices
   P     # Take the product of those two values
    d    # And check that it's non-negative / >=0 (thus no score is decreasing)
         # (after which the result is output implicitly)

05AB1E, 15 14 bytes

εDO5÷FR]{øÆ1‹P

Try it online or verify all test cases.

A minor alternative: FR]{øÆ1‹P could be ._}{R`-dß for the same byte-count:

Try it online or verify all test cases.

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 {       # Sort the list of pairs from lowest to highest
  ø      # Zip/transpose; swapping rows/columns
   Æ     # Reduce each inner pair by subtracting
    1‹   # Check for both values if they're smaller than 1 (thus no score is decreasing)
      P  # And check if both are truthy by taking the product
         # (after which the result is output implicitly)

εDO5÷    # Same as above
     ._  #  Rotate the pair that many times towards the left
         #  (the pair is reversed for odd sums; and remains unchanged for even sums)
}{R      # After the map: sort the list of pairs from highest to lowest
   `     # Pop and push both pairs separated to the stack
    -    # Subtract the values of the pairs from one another at the same indices
     d   # Check for each value if it's non-negative / >=0 (thus no score is decreasing)
      ß  # And check if both are truthy by taking the minimum
         # (after which the result is output implicitly)

05AB1E, 15 14 13 bytes

εDO5÷FR]{`-Pd

-1 byte thanks to @Neil.

Try it online or verify all test cases.

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 {       # Sort the list of pairs from lowest to highest
  `      # Pop and push both pairs separated to the stack
   -     # Subtract the values of the pairs from one another at the same indices
    P    # Take the product of those two values
     d   # And check that it's non-negative / >=0 (thus no score is decreasing)
         # (after which the result is output implicitly)

05AB1E, 15 14 bytes

εDO5÷FR]{øÆ1‹P

Try it online or verify all test cases.

Explanation:

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 {       # Sort the list of pairs from lowest to highest
  ø      # Zip/transpose; swapping rows/columns
   Æ     # Reduce each inner pair by subtracting
    1‹   # Check for both values if they're smaller than 1 (thus no score is decreasing)
      P  # And check if both are truthy
         # (after which the result is output implicitly)

05AB1E, 15 14 bytes

εDO5÷FR]{øÆ1‹P

Try it online or verify all test cases.

A minor alternative: FR]{øÆ1‹P could be ._}{R`-dß for the same byte-count:

Try it online or verify all test cases.

Explanation:

ε        # Map both pairs in the (implicit) input to:
 D       #  Duplicate the pair
  O      #  Pop this duplicate and calculate its sum
   5÷    #  Integer-divide it by 5
     F   #  Loop that many times:
      R  #   Reverse the pair every iteration
         #   (the pair is reversed for odd sums; and remains unchanged for even sums)
]        # Close both the loop and map
         # (all pairs are now in the order [A,B])
 {       # Sort the list of pairs from lowest to highest
  ø      # Zip/transpose; swapping rows/columns
   Æ     # Reduce each inner pair by subtracting
    1‹   # Check for both values if they're smaller than 1 (thus no score is decreasing)
      P  # And check if both are truthy by taking the product
         # (after which the result is output implicitly)

εDO5÷    # Same as above
     ._  #  Rotate the pair that many times towards the left
         #  (the pair is reversed for odd sums; and remains unchanged for even sums)
}{R      # After the map: sort the list of pairs from highest to lowest
   `     # Pop and push both pairs separated to the stack
    -    # Subtract the values of the pairs from one another at the same indices
     d   # Check for each value if it's non-negative / >=0 (thus no score is decreasing)
      ß  # And check if both are truthy by taking the minimum
         # (after which the result is output implicitly)