Prove inequality 3 ^ n ≥ n ^ 3 (n ≥ 3) RT It's better not to use derivatives N is a positive integer

Prove inequality 3 ^ n ≥ n ^ 3 (n ≥ 3) RT It's better not to use derivatives N is a positive integer

Induction proves that:
When n = 1, 3 ^ n = 3, n ^ 3 = 1, 3 ≥ 1, so 3 ^ n ≥ n ^ 3 holds;
When n = 2, 3 ^ n = 9, n ^ 3 = 8, 9 ≥ 9, so 3 ^ n ≥ n ^ 3 holds;
When n = 3, 3 ^ n = 27, n ^ 3 = 27, 27 ≥ 27, so 3 ^ n ≥ n ^ 3 holds;
Suppose that when n = K (K ≥ 3), the original inequality holds, that is, 3 ^ k ≥ K ^ 3,
Then, when n = K + 1, 3 ^ (K + 1) = 3 × 3 ^ k ≥ 3K ^ 3 = (3 * k) ^ 3 under the third root sign
When k ≥ 3, 3 * k - (K + 1) = (3-1) * k-1, 3-1) * k-1 ≥ 0
So (3 * k) ^ 3 under the third root) ≥ (K + 1) ^ 3, so 3 ^ (K + 1) ≥ (3 * k) ^ 3 under the third root) ≥ (K + 1) ^ 3
That is, when n = K + 1, the original inequality holds
So the original inequality holds for any n ∈ n +, that is to say, the original inequality is proved