exponential difference inequality Im asked to prove the inequality when $0\leq a<b$ and $x>0$:
$$
a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a)
$$
So far I have seen that obviously:
$$a^x(b-a)<b^x(b-a)$$
and that
$$b^{x+1}-a^{x+1} = (b-a)(b^x+b^{x-1}a+...+ba^{x-1}+a^x) > a^x(b-a)$$
This has done no good for me yet but it seems related to it.
$$$$
I was thinking it may have to do with $a<{a+b\over{2}}<b$ but I cant see how. If you think a hint may help me out I would prefer that over a straight solution. But any help is appreciated thank you.
 A: One approach is to invoke the mean-value theorem.  Let $f(z)=z^{x+1}$ for $x>0$ and $0<a\le z\le b$.  Then, there exists a number $\xi\in(a,b)$ such that
$$\frac{b^{x+1}-a^{x+1}}{b-a}=(x+1)\xi^{x} \tag 1$$
Rearranging $(1)$ reveals 
$$\frac{b^{x+1}-a^{x+1}}{x+1}=\xi^{x}(b-a) \tag 2$$
Since $a<\xi<b$, then $a^x<\xi^x<b^x$, we have
$$a^{x}(b-a)<\frac{b^{x+1}-a^{x+1}}{x+1}<b^{x}(b-a)$$
as was to be shown!
A: other  simple solution
since
$$a^x<t^x<b^x,t\in (a,b),x>0$$
so
$$\int_{a}^{b}a^x dt<\int_{a}^{b}t^x dt<\int_{a}^{b}b^xdt $$
so
$$(b-a)a^x<\dfrac{b^{x+1}-a^{x+1}}{x+1}<(b-a)b^x$$
A: also use your idea:
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x>a^x+a^x+\cdots+a^x=(x+1)a^x$$
and 
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x<b^x+b^x+\cdots+b^x=(x+1)b^x$$
A: Take user math110's idea: 
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x>a^x+a^x+...+a^x=(x+1)a^x$$
and
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x<b^x+b^x+\cdots+b^x=(x+1)b^x$$
From this and since $0\leq a<b$
$$
(x+1)a^x<(b^x+b^{x-1}a+...+ba^{x-1}+a^x)<(x+1)b^x
$$
Divide by $x+1$ and we get,
$$
a^x<{b^x+b^{x-1}a+...+ba^{x-1}+a^x\over{x+1}}<b^x
$$
Multiply by $b-a$, the middle term becomes,
$$
{(b-a)(b^x+b^{x-1}a+...+ba^{x-1}+a^x)\over{x+1}}={b^{x+1}-a^{x+1}\over{x+1}}
$$
and we have,
$$
a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a)
$$
