Proving $b-1\geq n(b^{1/n}-1)$

Question

Question

Prove $b-1\geq n(b^{1/n}-1)$ from $b^n-1\geq n(b-1)$ where $n$ is an integer and $b>1,y>0$.

Attempt

Why does putting $n = 1/n'$ doesn't provide the required expression?

$$b^n-1\geq n(b-1) $$ $$ b^{1/n'}-1\geq 1/n'(b-1)$$ $$ n'(b^{1/n'}-1)\geq (b-1)$$ $$ n(b^{1/n}-1)\geq (b-1)$$

P.S. I am aware of how to solve the question (Substitute $ b=b^{1/n'} $). Just interested to know why this method doesnt work.

If the argument is $1/n'$ is not an integer for integer $n'$ then why does the substitution $ b=b^{1/n'} $ works?

Answer 1

Because $n' < 1$ when $n > 1$. You can't apply the same inequality for $n$ and substitute with $n'$ if they are not both greater or both less than $1$.

The inequality sign is reversed in the Bernoulli inequality when the exponent is less than $1$:

$$b^n \geqslant 1 + n(b-1), \\ b^{1/n} \leqslant 1 + \frac{1}{n} (b-1).$$

Also the exponent can be any real positive real number.

Answer 2

Let $b^{\frac{1}{n}}=t$. Thus, $$b-1=t^n-1=(t-1)(t^{n-1}+...+1)\geq(t-1)n=n\left(b^{\frac{1}{n}}-1\right).$$