Prove by induction the inequality (1 + x)n ≥ 1 + nx, whenever x is p

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

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

Answer»

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.