Bernoulli's Inequality/Proof 1

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Let $x \in \R$ be a real number such that $x > -1$.

Let $n \in \Z_{\ge 0}$ be a positive integer.


$\paren {1 + x}^n \ge 1 + n x$


Proof by induction:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\left({1 + x}\right)^n \ge 1 + nx$

Basis for the Induction

$P \left({0}\right)$ is the case:

$\left({1 + x}\right)^0 \ge 1$

so $P \left({0}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\left({1 + x}\right)^k \ge 1 + kx$

We need to show that:

$\left({1 + x}\right)^{k+1} \ge 1 + \left({k + 1}\right) x$

Induction Step

This is our induction step:

\(\displaystyle \left({1 + x}\right)^{k+1}\) \(=\) \(\displaystyle \left({1 + x}\right)^k \left({1 + x}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle \left({1 + kx}\right) \left({1 + x}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle 1 + \left({k + 1}\right) x + k x^2\)
\(\displaystyle \) \(\ge\) \(\displaystyle 1 + \left({k + 1}\right) x\)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


Source of Name

This entry was named for Jacob Bernoulli.