Question

Find and prove an inequality relating 100n and n^{3} .

Answer 1

An inequality relating 100n and n³ is 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

What is inequality?

An inequality is comparison of  two values, showing if one is less than, greater than, or simply not equal to another value.

Since 100n and n³ for n = 1, 2, 3, . . . 9, 10, 11 are 100, 200, 300, . . . 900, 1000, 1100 and 1, 8, 27, . . . 729, 1000, 1331 respectively.

Therefore, an inequality relating 100n and n³ will be 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

Induction hypothesis:

Suppose 100n ≤ k³ for some positive integer k ≥ 10.

We need to show that 100( k + 1 ) ≤ ( k + 1 )³ = k³ + 3k² +3k + 1.

Note 100( k + 1 ) = 100k + 100 ≤ k³ + 100

                                                 ≤ k³ + 3k²    (∵ k ≥ 10 )

                                                 ≤ k³ + 3k² + 3k

                                                 ≤ k³ + 3k²+3k + 1

So 100( k + 1 ) ≤ ( k + 1 )³, which is true.

Hence by the principle of mathematical induction, 100n ≤ k³ for every integer  k ≥ 10.

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