Several inequalities

Let $m,n\in \Bbb N, x\in \Bbb R, x\geq 0, x\neq 1$. Prove the followings: if $m>n$, then $\dfrac{x^m-1}{m}>\dfrac{x^n-1}{n}$; if $m<n$, then $\dfrac{x^m-1}{m}<\dfrac{x^n-1}{n}$.

Proof:
Consider the case when $m>n$. The other case is similar. It suffices to show that $n(x^m-1)-m(x^n-1)> 0$ for $x>1$. $$\begin{align}n(x^m-1)-m(x^n-1)&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+1)-m(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+x^n)+n(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&\quad-m(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&=n(x-1)(x^{m-1}+x^{m-2}+\cdots+x^n)\\&\quad-(m-n)(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\\&> (x-1)[n\underbrace{(x^n+\cdots+x^n)}_{m-n\; terms}-(m-n)\underbrace{(x^n+\cdots+x^n)}_{n\; terms}]\qquad(*)\\&=(x-1)x^n[n(m-n)-(m-n)n]\\&=0\qquad \Box\end{align}$$ In $(*)$, we have used the facts that $x^{m-i}\geq x^n$ when $i=1,\cdots,m-n$ and $-x^{n-j}\geq -x^n$ where $j=1,\cdots,n$. The case for $0\leq x<1$ also holds because we have $-x^{m-i}\geq -x^n$ when $i=1,\cdots,m-n$ and $-x^{n-j}\leq -x^n$ where $j=1,\cdots,n$.

Remark: This result implies that the sequence of functions $f_n(x)=\dfrac{x^n-1}{n}$ is increasing when $x$ is fixed and $x>1$. The term $-1$ there is crucial. Recall the taylor series of $e^x$ and exponential functions: $$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\\ 2^x=e^{(\ln 2)x}=1+(\ln 2)x+\dfrac{((\ln 2)x)^2}{2!}+\cdots.$$ Then $\dfrac{e^x-1}{x}=1+\dfrac{x}{2}+\dfrac{x^2}{6}+\dfrac{x^3}{24}+\cdots$, whereas $\dfrac{e^x}{x}$ or $\dfrac{e^x+5}{x}$ contains the expression $\dfrac{1}{x}$, which means there will be asymptotes.

We can use the proved inequalities to prove some other inequalities.

Let $a,b\in \Bbb R, a>0, b>0, a\neq b, r\in \Bbb Q$ Then $$rb^{r-1}(a-b)<a^r-b^r<ra^{r-1}(a-b)\quad r>1\\ ra^{r-1}(a-b)<a^r-b^r<rb^{r-1}(a-b)\quad 0<r<1.$$ Proof:
When we divide the inequalities by $b^r$, we will see that they resemble the inequalities that we have proved. We just prove the case when $r>1$. Let $r=\dfrac{m}{n}$, where $m,n\in \Bbb N$. $r>1\Rightarrow m>n$. Substituting $x=(\dfrac{a}{b})^{1/n}$ gives $$\dfrac{(\dfrac{a}{b})^{m/n}-1}{m}>\dfrac{(\dfrac{a}{b})^{m/n}-1}{n}\\ \dfrac{a^r-b^r}{mb^r}>\dfrac{a-b}{bn}\\ a^r-b^r>rb^{r-1}(a-b).$$ Similarly, substituting $x=(\dfrac{b}{a})^{1/n}$ yields $$\dfrac{(\dfrac{b}{a})^{m/n}-1}{m}>\dfrac{(\dfrac{a}{b})^{n/n}-1}{n}\\ \dfrac{b^r-a^r}{ma^r}>\dfrac{b-a}{an}\\ a^r-b^r<ra^{r-1}(a-b).\qquad \Box$$