<s>[INST] How to prove $1Let $a,b>0$, $a\neq b$, $n\in\mathbb N$, $n>1$, and $a^n-b^n=a^{n+1}-b^{n+1}$.
How can we prove that
$1<a+b<\dfrac{2n}{n+1}$
Thank you  everyone. I have proven it. My method:
$a+b-1=a+b-\dfrac{a^{n+1}-b^{n+1}}{a^n-b^n}=\dfrac{ab(a^{n-1}-b^{n-1})}{a^n-b^n}>0$
on the other hand
$$a+b-\dfrac{2n}{n+1}=a+b-\dfrac{2n}{n+1}\cdot\dfrac{a^{n+1}-b^{n+1}}{a^n-b^n}=\dfrac{(n+1)ab(a^{n-1}-b^{n-1})-(n-1)(a^{n+1}-b^{n+1})}{(n+1)(a^n-b^n)}$$
so 
$$\Longleftrightarrow ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1}),n>1$$
let $a>b>0,t=\dfrac{a}{b}$
$$\Longleftrightarrow (n+1)t(t^{n-1}-1)<(n-1)(t^{n+1}-1),t>1 $$
 follwing is very easy.
so we let $f(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-(n-1),x>1$
then we $f'(x)=(n-1)(n+1)x^n-n(n+1)x^{n-1}+(n+1),x>1$
$f''(x)=n(n-1)(n+1)x^{n-1}-n(n+1)(n-1)x^{n-2}=(n-1)n(n+1)x^{n-2}(x-1)>0,x>1$ 
so 
$f'(x)>f'(1)=0$
then 
$f(x)>f(1)=0$
so 
$$ (n+1)t(t^{n-1}-1)<(n-1)(t^{n+1}-1),t>1 $$
I think there are other methods?
 [/INST] Firstly, a point of caution. In your solution, you had a step which went:
\begin{align}
& \frac{(n+1)ab(a^{n-1}-b^{n-1})-(n-1)(a^{n+1}-b^{n+1})}{(n+1)(a^n-b^n)}<0 \\
& \Longleftrightarrow ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1})
\end{align}
Note that in order to do this, you must first WLOG assume $a>b$, so that the denominator is positive and you don't need to worry about it.
Now, here's an alternative that continues from the step above. Since $a>b$, we divide by $a-b$ to get:
\begin{align}
& ab(a^{n-1}-b^{n-1})(n+1)<(n-1)(a^{n+1}-b^{n+1}) \\
& \Longleftrightarrow (n+1)(a^{n-1}b+a^{n-2}b^2+ \ldots +ab^{n-1})<(n-1)(a^{n}+a^{n-1}b+ \ldots +b^{n}) \\
& \Longleftrightarrow (n-1)(a^n+b^n)>2(a^{n-1}b+a^{n-2}b^2+ \ldots +ab^{n-1})
\end{align}
By AM-GM, 
$$a^n+b^n=\frac{(n-i)a^n+ib^n}{n}+\frac{ia^n+(n-i)b^n}{n} \geq a^{n-i}b^i+a^ib^{n-i}$$
Summing over $i=1, 2, \ldots . n-1$ gives the desired inequality. The inequality is strict because $a \not =b$.
</s>