Lesson 6

After proving two formulas (that is, equalities) by the method of mathematical induction, we turn to proving two frequently used inequalities.

Theorem (Bernoulli's inequality). For any \(x \in \mathbb{R}_{\ge -1}\) and \(n \in \mathbb{Z}_{\ge 1}\), the inequality \[(1+x)^{n} \ge 1+xn \ \] holds.

Proof. The base case \(n=1\) is true. Inductive step \(n \to (n+1)\): \[\begin{align*}(1+x)^{n+1}&=(1+x)^{n} \cdot (1+x)\\&\ge (1+xn) \cdot (1+x)&\text{(hypothesis and \((1+x) \ge 0\))}\\&=1+x(n+1)+x^{2}n \\&\ge 1+x(n+1) \ .&\text{(\(x^{2}n \ge 0\))}\end{align*}\]◼