# Generalization of Bernoulli's Inequality.

When you're first taking an Introduction to Proofs class, you eventually learn about induction. One of the first induction proofs I did was to prove for $n \geq 1$ that $(1+x)^n \geq 1+nx$. You first prove the base case, $n=1$, which yields $(1+x)^1 \geq 1+1x$, which is true. Then you assume that the inequality holds for for the arbitrary $n$ case. Then you try to prove that the inequality holds for $n+1$, i.e. $(1+x)^(n+1) \geq 1+nx$ as well. Using our assumption that $(1+x)^n \geq 1+nx$, we can multiply both sides by $1+x$, which yields $(1+x)*(1+x)^n \geq (1+nx)*(1+x).$ Since $1+nx^2+nx +x \geq 1+(n+1)x$, the inequality is proven. Now what I’m trying to do is prove a generalization of Bernoulli’s inequality, i.e. the following.

“Find the least possible $\alpha < 0$ so that $\forall x \geq \alpha$ and $n \in \mathbb{N}$, that $(1+x)^n \geq 1+nx$. Prove the inequality as well as prove that your $\alpha$ is best possible.”

What I’m trying to figure out is what is more special about this question rather than the typical inductive Bernoulli’s inequality proof? What is the purpose of introducing this alpha restricted to less than zero? There must be a reason for it. I read a few different textbooks and they all say that it (it referring to Bernoulli's inequality) works for $x > -1$, so I’m assuming that the alpha would also be -1 because then for any $x \in \mathbb{R}$ such that $x \geq \alpha$, the normal Bernoulli’s inequality holds. Am I completely going off the deep end? Any input would be very much appreciated.

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Does the inequality hold when $x=-1$ ? – Ju'x Dec 6 '12 at 22:55

## 2 Answers

We find the answer, using basic calculus. The magic $\alpha$ is not $-1$. Consider the function $f(x)=(1+x)^n -(1+nx)$ for $n \gt 1$. Then $$f'(x)=n((1+x)^{n-1}-1).$$ For $n\gt 1$ and even, the derivative vanishes only at $x=0$. For $n\gt 1$ and odd, the derivative vanishes at $x=-2$ and $x=0$.

The only interesting case is $n$ odd. By the usual increasing/dcreasing analysis, we find that $f(x)\ge 0$ when $x\ge -2$. For any $w\lt -2$, by taking a suitably large $n$, we can make $f(w)\lt 0$.

The part about $f(x)\ge 0$ for $x\ge -2$ can be pushed through by induction.

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So the critical points of f(x) occur at x = 0 for all n and x = -2 for n odd. So x = -2 is not a critical point for n even; therefore, how can x = -2 be a minimum for all n?

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