1. ## Bernoulli's inequality consequence

Hello community. I'm asked to show that if 1 < uk then (u-1/k)^n is greater or equal to u^n - (u^(n-1))*n/k. I'm supposed to use Bernoulli's inequality. I'd appreciate any hint. Please don't solve it for me

2. ## Re: Bernoulli's inequality consequence

Is $n$ a positive integer? Second, state what the Bernoulli inequality is (for the benefit of our readers).

3. ## Re: Bernoulli's inequality consequence

Ohh sure, I'm sorry. Bernoulli's inequality states that if -1 < x then (x+1)^m is greater or equal to 1+m*x. This is valid for all natural number m. And yes, n is a positive integer.

4. ## Re: Bernoulli's inequality consequence

So given

$x > -1$ then $(1+x)^n \ge 1 + n x$

you want to show if

$uk > 1$ then $\left( u - \dfrac{1}{k}\right)^n \ge u^n - \dfrac{u^{n-1}n}{k}$.

Comparing the first part of the inequalities, have you thought of letting $x = \dfrac{-1}{uk}$?

5. ## Re: Bernoulli's inequality consequence

Yes, multiplying by u^n at both sides of the inequality brings the result I want but, what if u^n is negative?

6. ## Re: Bernoulli's inequality consequence

Originally Posted by Jester
So given
$x > -1$ then $\color{red}(1+x)^n > 1 + n x$
You ought to at least get the statement right. It should be $(1+x)^n \ge 1 + n x$ . Otherwise $n=1$ is false.

The standard prove follows easily by proof by induction.

7. ## Re: Bernoulli's inequality consequence

Originally Posted by Plato
You ought to at least get the statement right. It should be $(1+x)^n \ge 1 + n x$ . Otherwise $n=1$ is false.

The standard prove follows easily by proof by induction.
You are right Plato, but the issue here isn't that inequality but the other one. Jester has given me a hint that works perfectly when u^n is positive. I don't know how to treat the case when u^n is negative though.

8. ## Re: Bernoulli's inequality consequence

I have found a counter-example, taking u = -7, k = -1/3 and n = 3. Apparently the author of the book from which I took the exercise meant 1 < u and 1 < k or maybe 1 < uk with the condition that both be positive. I had to go back to the text where he proves the existence of the positive nth root of c > 0, there he uses that inequality (which was giving me a headache already) in a crucial step but he's only working with positive numbers, then he asks one to prove it as an exercise later. So if we attach to standard notation, that meaning:"1 < uk" means the product of u and k is greater than one, he's basically telling you to prove something false.