Question

For all n ∈ N, by using principle of mathematical induction, then prove that: (1 + x)ⁿ ≥ 1 + nx, x > 0

Answer 1

Let P(n) : (1 + x)ⁿ ≥ (1 + nx), x > 0

when n = 1 then P(n) is true

⇒ (1 + x) ≥ (1 + x) is true for x > -1 …(i)

For n = k

P(m) : (1 + x)^m ≥ (1 + mx), x > 0 is true …(ii)

Now, we have to prove that P(m + 1)

We have that

(1 + x)^m+1 = (1 + x)^m (1 + x)

Given x > 0, So, (1 + x) > 0

Hence, using

(1 + x)^m ≥ (1 + mx)

we get (1 + x)^m+1 ≥ (1 + mx)(1 + x)

i.e., (1 + x)^m+1 ≥ (1 + x + mx + mx²) …..(iii)

Since, m is a natural number and x² ≥ 0 is such that

mx² ≥ 0

So, (1 + x + mx + mx²) ≥ (1 + x + mx)

⇒ (1 + x)^m+1 ≥ (1 + x + mx)

⇒ (1 + x)^m+1 ≥ [1 + (1 + m)x]

Since, statement (ii) is true.

Hence, by the principle of mathematical induction

P(n) is true for each natural number n ∈ N.

Hence Proved.