Bernoulli's Inequality via Taylor's Theorem

Worked Solution

How to Think About It: The right-hand side $1 + nx$ is the tangent line to $f(x) = (1 + x)^n$ at $x = 0$. Taylor's theorem writes $f$ as that tangent line plus a second-order remainder, so the inequality holds as soon as the remainder is nonnegative, which is a sign check on $f''$.

Approach: Expand $f(x) = (1+x)^n$ to first order about $0$ with the Lagrange form of the remainder, then show the remainder term is $\ge 0$ whenever $x > -1$. Confirm with induction.

Formal Solution:

*Step 1 -- Taylor expansion with Lagrange remainder.* Let $f(x) = (1 + x)^n$, which is infinitely differentiable on $(-1, \infty)$. For any $x > -1$ there exists $\xi$ strictly between $0$ and $x$ (so $\xi > -1$) such that $$f(x) = f(0) + f'(0)\,x + \frac{f''(\xi)}{2}\,x^{2}.$$

*Step 2 -- Compute the terms.* $$f(0) = 1, \qquad f'(x) = n(1+x)^{n-1} \Rightarrow f'(0) = n, \qquad f''(\xi) = n(n-1)(1+\xi)^{n-2}.$$

*Step 3 -- Sign of the remainder.* Since $n \ge 2$, $n(n-1) > 0$; since $\xi > -1$, $(1 + \xi)^{n-2} > 0$ (for $n = 2$ it equals $1$). Therefore $$\frac{f''(\xi)}{2}x^{2} = \frac{n(n-1)}{2}(1+\xi)^{n-2}x^{2} \ge 0,$$ and consequently $(1 + x)^n = 1 + nx + (\text{nonnegative}) \ge 1 + nx$.

*Step 4 -- Equality.* The remainder is zero exactly when $x^2 = 0$, so for $n \ge 2$ equality holds only at $x = 0$. For $x > -1$, $x \ne 0$, the inequality is strict.

*Step 5 -- Induction check (alternative proof).* Base case $n = 2$: $(1+x)^2 = 1 + 2x + x^2 \ge 1 + 2x$. Inductive step: if $(1+x)^k \ge 1 + kx$, multiply by $1 + x > 0$: $$(1+x)^{k+1} \ge (1 + kx)(1 + x) = 1 + (k+1)x + kx^{2} \ge 1 + (k+1)x.$$ The hypothesis $x > -1$ is used precisely to keep $1 + x$ positive so the inequality direction is preserved.

Answer: For all $x > -1$ and integers $n \ge 2$, $(1+x)^n \ge 1 + nx$, because the Taylor remainder $\frac{n(n-1)}{2}(1+\xi)^{n-2}x^2$ is nonnegative; equality holds only at $x = 0$.