Question

Prove by induction the inequality (1 + x)ⁿ ≥ 1 + nx, whenever x is positive and n is a positive integer.

Answer 1

P(n) : (1 + xⁿ) ≥ 1 + nx 

P(1) : (1 + x)¹ ≥ 1 + x 

⇒ 1 + x ≥ 1 + x, which is true. 

Hence, P(1) is true. 

Let P(k) be true 

(i.e.) (1 + x)^k ≥ 1 + kx 

We have to prove that P(k + 1) is true. 

(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x 

Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true] 

Multiplying both sides by (1 + x), we get 

(1 + x)^k (1 + x) ≥ (1 + kx)(1 + x) 

⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx²

⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ... (1) 

Now, 1 + (k + 1) x + kx₂ ≥ 1 + (k + 1)x … (2) [∵ kx > 0] 

From (1) and (2), we get 

(1 + x)^{k + 1} ≥ 1 + (k + 1)x 

∴ P(k + 1) is true if P(k) is true. 

Hence, by the principle of mathematical induction,

P(n) is true for all values, of n.