inequality with power function

Question

Hello. Let $0<b\leq1$ and $n \in \mathbb{N}$ with $n\geq1$, does this inequality hold?
$(n-1)^b \leq n^b-1$

Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)

Answer 1

The inequality is actually the other way around. You can consider the function $$f(x)=(x-1)^b-x^b+1$$ well-defined for every $x\geq 1$. Then $$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$ and since $x\geq 1$ we have $0\leq x-1\leq x$ so that $f'(x)\geq 0$ for all $x\geq 1$ (here we use $b-1\leq 0$) Hence $f$ is increasing. Moreover $$f(1)=0$$ so that $f(x)\geq 0$ for every $x\geq 1$ that is $$(x-1)^b\geq x^b-1$$ for all $x\geq 1$, in particular if $x\in\mathbb{N}$.