Q1.
Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations? [Hint: Consider the direct lengths in the increasing order.]
Answer
Yes. Put the three lengths in increasing order — the two smaller ones pass automatically.
Let a ≤ b ≤ c
a ≤ c < b + c → a < b + c ✓ (no calculation needed)
b ≤ c < a + c → b < a + c ✓ (no calculation needed)
Only c < a + b has to be checked
a ≤ c < b + c → a < b + c ✓ (no calculation needed)
b ≤ c < a + c → b < a + c ✓ (no calculation needed)
Only c < a + b has to be checked
For example, in 6, 11, 14 you can say at once that 6 and 11 pass; only 14 vs 6 + 11 = 17 needs a look.
Why it happens: Comparing a small number with a sum that already contains the largest number is no contest. The only tight comparison is the one where the largest number stands alone on the left.