if $x_{1},x_{2} $ is $nx-x^n=a$ two roots, show that $|x_{1}-x_{2}|<\dfrac{a}{1-n}+2$ 
Assmue that $n$ be postive integers,and $a$is real number,and the equation $$nx-x^n=a$$ has postive real roots $x_{1},x_{2}$,show that
  $$|x_{1}-x_{2}|<\dfrac{a}{1-n}+2$$

By condition, I showed that
$$(x_{1}-x_{2})[n-(x^{n-1}_{1}+x^{n-2}_{1}x_{2}+\cdots+x^{n-1}_{2})]=0$$
But I don't know what to do then. Please help, Thanks.
 A: Let's study the function $f(x)=nx-x^n$ for $x>0$. $f'(x)=n(1-x^{n-1})$ so there is a global maximum in $1$. The function is monotonic increasing in $[0,1]$ and decreasing in $[1,\infty]$. The function attains twice the same value $a$ if and only if $a\in [0,n-1]$, that is $x\in [0,\sqrt[n-1]{n}]$. So for any such $x_1,x_2$ we have in fact $$|x_1-x_2|\leq\sqrt[n-1]{n}\leq 2\leq \frac{a}{n-1}+2.$$
Equality case: we must have $a=0$ and $2^{n-1}=n$, that is $n=1,2$. For $n=1$ $f(x)=0$ identically (and the text doesn't really make sense), so we might want to suppose $n\geq 2$ from the beginning, and for $n=2$ we have $f(x)=2x-x^2$: here $x_1=0$ and $x_2=1$ give us equality. In all other cases we have strict inequality.
A: Let $f(x)=nx-x^n-a$. Then by Rolle's theorem we get, $\exists c\in [x_1,x_2]$ such that $f'(c)=0\implies c=1$ assuming $n>1$. Then, $\exists \delta>0$ such that $x_1=1-\delta, x_2=1+\delta$. Now, $$nx_2-x_2^n=a\implies n(1+\delta)-(1+\delta)^n=a\\\implies a=(1+\delta)(n-(1+\delta)^{n-1})\\\implies a<(1+\delta)(n-(1+(n-1)\delta))\\\implies a< (n-1)(1-\delta^2)\\\implies 2(\delta-1)<(\delta^2-1)<\frac{a}{1-n}\\\implies |x_1-x_2|=2\delta<\frac{a}{1-n}+2 $$
