3
$\begingroup$

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 $?

$\endgroup$
2
$\begingroup$

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 ($t^2 > 0$, $n-1 > 0$).

$\endgroup$
  • 2
    $\begingroup$ Why $(1+t)^{n-1} \leq (1+(n-1)t)$? The Bernoulli's inequality states quite the opposite... $\endgroup$ – CMTV Jan 9 '20 at 10:14
  • 1
    $\begingroup$ What a stupid mistake. I edited. $\endgroup$ – Mindlack Jan 9 '20 at 10:39
  • $\begingroup$ 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 $n =2$. $\endgroup$ – CMTV Jan 9 '20 at 10:46
  • 1
    $\begingroup$ No: with $n=2$ you still get $t^2 \leq 0$ which implies $t=0$. $\endgroup$ – Mindlack Jan 9 '20 at 10:55
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.