If $0<b<a<1$ with $a,b$ close to $1$
Then, is $(a^r-b^r)-(a-b)$ alwasys a negative for any $r\in[1,\infty)$?
My try : consider, $f(r)=(a^r-b^r)-(a-b)$
If I show $f(r)$ is deacreasing on $[1,\infty)$, then $f(r)\lt f(1)=0$ for any $r\in(1,\infty)$
Now, $f'(r)=a^r\cdot log(a)-b^r\cdot log(b)\\\implies f'(r)=b^r\cdot log(b) \left(\left(\frac{a}{b}\right)^r log\left(\frac{a}{b}\right)-1\right)$
Now, here I have been stuck.
Here, also, one thing can be considered that $(a-b)$ can be taken sufficiently smaller to make above statement right.
Please help.
My main motivation is to show that there exist a sufficiently small $\delta\gt 0$ with $(a-b)\lt\delta$ such that $(a^r-b^r)-(a-b)$ alwasys a negative for any $r\in[2,\infty)$
The original statement is not correct, what follows is a proof that $a^r -b^r - (a-b)>0$ whenever $a>b>1$.
Note that $s:=\frac{a^r - b^r}{a-b}$ is the slope of the secant of $g(x)=x^r$ between $x=b$ and $x=a$. In particular, according to the mean value theorem, there is some $c\in[b, a]$ such that $s=rc^{r-1} \equiv g'(c)$. If $r>1$ and $1 < b < a$, we find that $s>1$.
Going back to the original question, we find for $0<b<a$: $$s>1\quad\Leftrightarrow\quad a^r - b^r > a-b.$$
Note that this requires that $a,b>1$, there is no upper bound to it.
The same proof helps illustrate why no conclusive statement is possible for $a,b<1$. Here, $0<c<1$, and both $s<1$ and $s>1$ is possible, depending on $r$. Hence $(a^r-b^r) - (a-b)$ will sometimes be positive, and sometimes negative. The intuition behind this is that the slope of $g(x)$ will be less than $1$ (and arbitrarily close to zero) near $x=0$, and approximately $r>1$ near $x=1$. Especially if $|a-b|$ "small", as mentioned by the user, the secant will be very close to the tangent.