Bernouilli's Inequality

# Bernouilli's Inequality

As an example of The Principle of Weak Mathematical Induction, we will now look at a very important inequality known as Bernouilli's inequality.

 Theorem (Bernouilli's Inequality): If $n$ is a nonnegative integer ($n = 0, 1, 2, ...$), then $(1 + x)^n ≥ 1 + nx$ for every $x ≥ -1$.
• Proof: This proof will be carried out by induction. Let $n$ be a nonnegative integer and let $P(n)$ be the statement that for all $x ≥ -1$, the inequality $(1 + x)^n ≥ 1 + nx$ holds.
• As our base step, let's check $n = 0$. Clearly $(1 + x)^0 = 1 ≥ 1$, so let's try $n = 1$ to be our base step just to be certain. We note that $(1 + x)^1 ≥ 1 + x$, that is $1 + x ≥ 1 + x$, which is true for all $x ≥ -1$.
• Now let $k$ be a non-negative integer and suppose that the statement $P(k) : (1 + x)^k ≥ 1 + kx$ for every $x ≥ -1$ is true. We want to show that the truth of $P(k)$ implies the truth of the statement $P(k+1): (1 + x)^{k+1} ≥ 1 + (k+1)x$ for every $x ≥ -1$.
• We apply our induction hypothesis and noting that since $k$ is a nonnegative, $kx^2 ≥ 0$, then:
(1)
\begin{align} \quad (1 + x)^{k+1} = (1 + x)(1 + x)^k \overset{IH} ≥ (1 + x)(1 + kx) = 1 + kx + x + kx^2 = 1 + (k + 1)x + kx^2 ≥ 1 + (k +1)x \end{align}
• Therefore for all nonnegative integers $n$, $(1 + x)^n ≥ 1 + nx$ for every $x ≥ -1$. $\blacksquare$

Bernouilli's inequality can understood graphically as well. For example, the following graph depicts the case where $n = 7$, and the lines $y_1 = (1 + x)^7$ (in red) and $y_2 = 1 + 7x$ (in teal). As we can clearly see, if $x ≥ -1$, then $y_1 ≥ y_2$.