Bernoulli's inequality

Example — Bernoulli's inequality

For $x>-1$ and $n\in\mathbb{N}$, $(1+x)^n \ge 1+nx.$ Base ($n=0$): $1\ge 1$. True.

Step: we multiply the hypothesis by $(1+x)>0$: $(1+x)^{k+1}=(1+x)^k(1+x)\ge(1+kx)(1+x)=1+(k+1)x+kx^2\ge 1+(k+1)x,$ because $kx^2\ge 0$. ∎

It will be used in Year 5 to prove that $\left(1+\dfrac{1}{n}\right)^n$ is bounded.

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Topics: Set theory
Concepts: Principle of induction
Methods: Induction
Skills: Proving
People: Jacob Bernoulli