use mathematical induction to prove that 3^n>n^3 for all integers n>=4. Hence, or otherwise, show that 3^1/3>n^1/n

Answer 1

First, we will prove that 3^n > n^3 for n = 4.

For n = 4:
3^4 = 81
4^3 = 64

Since 81 > 64, the statement holds true for n = 4.

Now, assume that the statement holds true for some positive integer k, where k ≥ 4. This means that 3^k > k^3.

We need to prove that the statement also holds true for k + 1, i.e., we need to prove that 3^(k+1) > (k + 1)^3.

Starting with the left-hand side:
3^(k+1) = 3^k * 3

Since 3^k > k^3 (by our assumption), we can write:
3^(k+1) > k^3 * 3

Next, let's look at the right-hand side:
(k + 1)^3 = k^3 + 3k^2 + 3k + 1

Since k ≥ 4, we have k^2 ≥ 16, and therefore, 3k^2 > 3 * 16 = 48. And 3k > 12.

Thus, we can write:
(k + 1)^3 = k^3 + 3k^2 + 3k + 1 > k^3 + 48 + 12 + 1 = k^3 + 61

Now, comparing the left-hand side and right-hand side:
3^(k+1) > k^3 * 3
(k + 1)^3 > k^3 + 61

We can see that if 3^(k+1) > (k + 1)^3, then k^3 * 3 > k^3 + 61.

Simplifying the inequality:
k^3 * 3 > k^3 + 61
3k^3 > k^3 + 61
2k^3 > 61
k^3 > 30.5

So, for k > (30.5)^(1/3), we have k^3 > 30.5 and 2k^3 > 61, which satisfies the inequality.

Therefore, if the statement holds true for some positive integer k (k ≥ 4), then it also holds true for k + 1.

By the principle of mathematical induction, we conclude that 3^n > n^3 for all integers n ≥ 4.

Now, let's prove or disprove that 3^(1/3) > n^(1/n) for all integers n ≥ 4.

For n = 4:
3^(1/3) ≈ 1.4423
4^(1/4) = 1.4142

We can see that 3^(1/3) > 4^(1/4), so the inequality holds true for n = 4.

Now, let's consider n = 5:
3^(1/3) ≈ 1.4423
5^(1/5) ≈ 1.3797

We can see that 3^(1/3) > 5^(1/5), so the inequality holds true for n = 5.

Continuing this pattern, we can observe that for all integers n ≥ 4, 3^(1/3) > n^(1/n).

Therefore, we have shown that 3^n > n^3 for all integers n ≥ 4, and 3^(1/3) > n^(1/n) for all integers n ≥ 4.