Alan'space

카테고리

Ideas

  • Mathematical induction

Claim

For n∈N0n \in \mathbb{N_0} and x≥−1x \geq -1, the following inequality holds:

(1+x)n≥1+nx(1+x)^n \geq 1 + nx

Proof

For n=0n=0, the inequality is trivially true:

(1+x)n=1≥1+nx=1(1+x)^n = 1 \geq 1 + nx = 1

Assume that the inequality holds for some n=k,k∈N0n=k, k \in \mathbb{N_0}. We need to show that it holds for n=k+1n=k+1.

Multiplying (1+x)(1+x) on both sides of the inequality:

(1+x)k+1≥(1+kx)(1+x)≥1+kx+x+x2≥1+(k+1)x+x2\begin{aligned} (1+x)^{k+1} &\geq (1 + kx)(1+x) \\ &\geq 1 + kx + x + x^2 \\ &\geq 1 + (k + 1)x + x^2 \end{aligned}

Now we have:

(1+x)k+1≥1+(k+1)x+x2≥1+(k+1)x(1+x)^{k+1} \geq 1 + (k+1)x + x^2 \geq 1 + (k+1)x

Which implies:

(1+x)k+1≥1+(k+1)x(1+x)^{k+1} \geq 1 + (k+1)x

We’ve shown if the inequality holds for n=kn=k, then it also holds for n=k+1n=k+1.

This completes the inductive step. By mathematical induction, the claim is true for all nn.