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I have seen different proofs of the Bernoulli inequality, but I am blocked in my solution:

$${(1+x)}^n\ge 1+nx \;\forall n\ge 0$$

The base case holds, I tried when n=0; now, I have put as an induction hypothesis that the theorem holds for n-1 and to proof the theorem for n. So, I have the following:

$$(1+x)^{n-1}\ge 1+(n-1)x$$ which is my induction hypothesis.

Now I wanted to proof that it holds por n, so I have the following:

$$(1+x)^{n-1}(1+x)\ge 1+(n-1)x(1+x)$$

$$\ge 1+nx+x^{2} (n-1)$$

but at this point I do not know how to continue the proof, I see that I get the a portion of tthe theorem that holds $$1+nx$$, and a part of the IH which is $$(n-1)x^{2}$$

How can I continue the proof from this part?

Thanks

Lila
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  • $$(n-1)x^2\ge0;\ldots$$ – DonAntonio Feb 01 '22 at 20:18
  • Deal with $n=0$ separately to $n \geq 1. $ Also, does the Bernoulli inequality your are trying to prove hold for all $x?$ Or only for $x\geq 0?$ Find out on wikipedia and be a bit more precise, but you are not that far away. – Adam Rubinson Feb 01 '22 at 20:23
  • Thank you @DonAntonio, but in most of the proofs they go from n to n+1, but I want to go from n-1 to n, – Lila Feb 01 '22 at 20:28
  • @Lila Yes, I know that...and in your development you got the above expression and that is my hint how to achieve what you want. – DonAntonio Feb 01 '22 at 21:44

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