Math 2.0

statement theorem

Bernoulli's inequality

Statement

Let xx be a real number with x≥−1x \ge -1, and let nn be a natural number. Then (1+x)n≥1+nx.(1 + x)^n \ge 1 + n x.

Remarks

When x≠0x \neq 0 and n≥2n \ge 2, the inequality is strict. For natural numbers nn, the inequality also holds when x≥−2x \ge -2. A proof by induction on nn multiplies (1+x)n≥1+nx(1 + x)^n \ge 1 + n x by 1+x≥01 + x \ge 0 and uses nx2≥0n x^2 \ge 0.

Explanations

L1 human

A small growth rate xx, applied nn times, gives at least nn times that growth. For example, 1.1101.1^{10} is at least 1+10⋅0.1=21 + 10 \cdot 0.1 = 2; its true value is about 2.592.59. The difference comes from the growth on the growth, which the right side ignores.

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Sources

  1. Web https://en.wikipedia.org/wiki/Bernoulli%27s_inequality, Integer exponent