The Bernoulli's inequality, stated by the Swiss mathematician Jacob Bernoulli in 1689, is of fundamental importance because it allows us to establish upper and lower bounds for exponential and polynomial functions.
Theorem. (Bernoulli's Inequality). Let \(x \in \mathbb{R}\) such that \(x \geq -1\). Then for every \(n \in \mathbb{N}\), the following inequality holds:
\[ (1 + x)^n \geq 1 + nx \]
Proof. We proceed by induction on the natural number \(n\).
Base case: For \(n = 0\), we have:
\[ (1 + x)^0 = 1 = 1 + 0 \cdot x \]
Thus, the base case is verified.
Inductive hypothesis: Assume the inequality is true for some \(k \in \mathbb{N}\), i.e.,
\[ (1 + x)^k \geq 1 + kx \]
Inductive step: We prove that the inequality holds for \(k + 1\). Multiply both sides of the inductive hypothesis by \((1 + x)\). This operation preserves the direction of the inequality since \(x \geq -1\) implies \((1 + x) \geq 0\). We obtain:
\[ \begin{align} (1 + x)^{k+1} & \geq (1 + kx)(1 + x) \\ & = 1 + x + kx + kx^2 \\ & = 1 + (k + 1)x + kx^2 \\ & \geq 1 + (k + 1)x \end{align} \]
The last inequality is justified by the fact that \(kx^2 \geq 0\) for every \(x \in \mathbb{R}\) and \(k \in \mathbb{N}\). By the principle of induction, the inequality is proven for every \(n \in \mathbb{N}\).
Observation. The condition \(x \geq -1\) is necessary to ensure that multiplication by \((1 + x)\) in the inductive step preserves the direction of the inequality.
Example. Bernoulli's inequality can be used to estimate the exponential function \(e^x\). Knowing that
\[ e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \]
we can apply the inequality and obtain a lower bound.
Bernoulli's inequality tells us that:
\[ \begin{align*} e^x & = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \\ & \geq \lim_{n \to \infty} \left(1 + n \cdot \frac{x}{n}\right) = 1 + x. \end{align*} \]
Therefore, \(e^x \geq 1 + x\). This result provides a simple and immediate estimate for the exponential function without having to perform complicated calculations.