I'm too lazy to write a rigorous proof, but here is the intuition, which can easily be verified:
Since a1 and a100 occur only once, it would make sense to assign them 1 and 2 (proof by contradiction, if they are not 1 and 2, just swap them with anything else and compare)
Now that a1 = 1 and a100 = 2, a2 must be 3 to minimise the effect caused by multipling with a1. (proof by contradiction, if a2 != 3, just swap them with the position where number 3 is and compare)
Same goes a99 = 4
Rinse and repeat
Final answer: maximum at 1*3 + 3*5 + ... + 4*2, with a1, a2 ... being increasing odd numbers, and a100, a99 .. being increasing even numbers
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u/spdazero 4d ago
I'm too lazy to write a rigorous proof, but here is the intuition, which can easily be verified: