If I have a function $f(x)=x^n$ and $x>0$, $n>1$, there are four positive values $0<c<a<b<d$ which are satisfied the condition $c=a-m$ and $d=b+m$, $m$ is a positive constant value. Here $g_1=\frac{f(b)-f(a)}{b-a}$ and $g_2=\frac{f(d)-f(c)}{d-c}$.
$g_1$ and $g_2$ whcih one is larger.
Let $n>2$, $f(x)=\frac{(b+x)^n-(a-x)^n}{b-a+2x},$ where $0<x<a.$
Thus, we need to prove that $f$ increases, for which we need to prove that $$f'(x)>0$$ or $$\frac{(n(b+x)^{n-1}+n(a-x)^{n-1})(b+x-(a-x))-2((b+x)^n-(a-x)^n)}{(b-a+2x)^2}>0$$ or $$(n-2)(b+x)^n-(n-2)(a-x)^n+n(a-x)^{n-1}(b+x)-n(b+x)^{n-1}(a-x)>0.$$ Now, let $b+x=t(a-x).$
Thus, $t>1$ and we need to prove that $g(t)>0,$ where $$g(t)=(n-2)t^n-(n-2)+nt-nt^{n-1}>0.$$ Indeed, by AM-GM $$g'(t)=n(n-2)t^{n-1}+n-n(n-1)t^{n-2}=n\left((n-2)t^{n-1}+1-(n-1)t^{n-2}\right)\geq$$ $$\geq n\left((n-1)\left(t^{(n-1)(n-2)}\cdot1\right)^{\frac{1}{n-1}}-(n-1)t^{n-2}\right)=0.$$ The equality does not occur because $t>1$.
Id est, $g(t)>g(1)=0$ and we are done in this case.
The case $1<n<2$ is the similar.
Can you end it now?