<s>[INST] 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.
$$$$
I was thinking it may have to do with $a<{a+b\over{2}}<b$ but I cant see how. If you think a hint may help me out I would prefer that over a straight solution. But any help is appreciated thank you.
 [/INST] Take user math110's idea: 
$$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)
$$
</s>