# Equality of Bernoulli's inequality

The simple form of Bernoulli's inequality is:

$$(1+x)^n \geq 1+xn \quad \text{ where } n\in\mathbb{N} \wedge \ x \geq -1$$

It is really easy to prove it via mathematical induction.

The only two options in which inequality becomes an equality are:

1. $$x=0$$
2. Any $$x$$ and $$n = 1$$

Now I wonder how do I prove that these two are the only options?

It would be better to prove that without using derivatives and so on...

Here is what I have tried:

Lets assume there is a special $$t\neq 0$$, $$t\geq -1$$ such as: $$(1+t)^n = 1+nt$$

Expanding the left part of an equation: $$(1+t)^n = 1 + nt + \binom{n}{2} t^2 + \ldots$$

This sum is obviously larger then $$1+nt$$ when $$t>0$$. So we have a contradiction.

How to get a contradiction when $$-1 \leqslant t < 0$$?

For any $$n \geq 2$$ we assume that $$(1+t)^n = 1+nt$$:

$$1+nt=(1+t)^n=(1+t)^{n-1}(1+t) \geq (1+(n-1)t)(1+t)=1+nt+(n-1)t^2 \\[15px] 1 + nt \geq 1 + nt + (n-1)t^2 \\[15px] (n-1)t^2 \leq 0$$

We get contradiction because this multiplication can be only positive ($$t^2 > 0$$, $$n-1 > 0$$).

• Why $(1+t)^{n-1} \leq (1+(n-1)t)$? The Bernoulli's inequality states quite the opposite... – CMTV Jan 9 '20 at 10:14
• What a stupid mistake. I edited. – Mindlack Jan 9 '20 at 10:39
• I see now. But I think $n$ must be strictly larger than 2 to get a contradiction here. But it is not a problem since we can manually get a contradiction for $n =1$ and $n =2$. – CMTV Jan 9 '20 at 10:46
• No: with $n=2$ you still get $t^2 \leq 0$ which implies $t=0$. – Mindlack Jan 9 '20 at 10:55

We can prove this via induction.
Let us fix $$t \in [-1, 0)$$. We show that if $$n \ge 2$$, then $$(1 + t)^n > 1 + nt$$.

Proof. Let us show the base case for $$n = 2$$.
$$(1 + t)^2 = 1 + 2t + t^2 > 1 + 2t$$, where the last inequality follows from the fact that $$t^2 > 0$$ as $$t \neq 0$$.

Now, let us assume that the statement is true for $$n = k \ge 2$$. We then have: $$(1+t)^{k+1} = (1+t)^k(1+t) \ge (1 + kt)(1 + t).$$ This $$\ge$$ is because $$1 + t \ge 0$$ and $$(1 + t)^k > 1 + kt$$. (Note that we need not have strict inequality here.)
Now, $$(1 + kt)(1 + t) = 1 + (k+1)t + kt^2 > 1 + (k+1)t.$$ Here, we have strict inequality as $$k > 0$$ and $$t^2 > 0$$.

Thus, we get that $$(1 + t)^{k+1} > 1 + (k+1)t$$.

By principle of mathematical induction, we are done.