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I will give 2 solutions
First solution
It will be easy if you know calculus
s − 1 ≤ ∫ 1 1 0 0 x − 1 / 3 d x ≤ s s − 1 ≤ ∫ 1 1 0 0 d x d 2 3 x 3 2 d x ≤ s s − 1 ≤ ∫ 1 1 0 0 d 2 3 x 3 2 ≤ s s − 1 ≤ 2 3 1 0 0 2 − 2 3 1 2 = 2 3 ( 1 0 0 2 − 1 ) = 3 0 . 8 1 6 5 2 0 4 . . . ≤ s s ≈ 3 1
Second solution
Through the coding environment in Brilliant: