If $a\ge 1$ and $0<r<1$ rational, then $a^r-1\le ra(a-1)$

Question

Let $a\ge1$ and $0<r<1$ be a rational number. I need to prove that $a^r-1\le ra(a-1)$.

Since $0<r<1$ we can put $r=\frac mn$ where $m<n$. Now, we have

$$a-1=(a^\frac 1n-1)(a^\frac{n-1}n+\cdots+a^\frac 1n+1)\ge(a^\frac 1n-1)(1+\cdots+1)=n(a^\frac 1n-1)$$

So, $a^\frac 1n-1\le\frac{a-1}n$ and if we replace $a$ by $a^m$, we get

$$a^r-1\le\frac{a^m-1}n=\frac{(a-1)(a^{m-1}+\cdots+a+1)}n$$

Could someone help me to continue from here? Thanks.

Answer 1

Hint: If we put $b=a^{\frac{1}{n}}$, then by the geometric series formula $$ \frac{a^{\frac{m}{n}}-1}{a^{\frac{1}{n}}-1} = \frac{b^m-1}{b-1} = \sum_{j=0}^{m-1} b^j = \sum_{j=0}^{m-1} a^{\frac{j}{n}} \leq \sum_{j=0}^{m-1} a = ma,$$ where the inequality follows since $\frac{j}{n} \leq 1$ as $j\leq m-1 \leq m < n$, and since $a\geq 1$. (Note that we should consider the case when $a=1$ separately here, but the inequality is trivial in that case.) Can you use your inequality $a^{\frac{1}{n}} - 1 \leq \frac{a-1}{n}$ to continue from here?