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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.

user3258845
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  • This is the standard inequality $$\frac{a^{r} - 1}{r} \geq \frac{a^{s} - 1}{s},,,\frac{1 - b^{r}}{r} \leq \frac{1 - b^{s}}{s}$$ where $0 < b < 1 < a$ and $r > s > 0$ are real numbers. In the current question we have $r = x +1, s = 1$ and role of $a$ is played by $b/a > 1$. See a proof of this at http://math.stackexchange.com/a/1491888/72031 – Paramanand Singh Dec 11 '15 at 06:08

4 Answers4

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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!

Mark Viola
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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 $$(b-a)a^x<\dfrac{b^{x+1}-a^{x+1}}{x+1}<(b-a)b^x$$

math110
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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 $$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x<b^x+b^x+\cdots+b^x=(x+1)b^x$$

math110
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  • Thank you figured it out. I will post proof in a minute – user3258845 Dec 11 '15 at 03:59
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    this is valid only when $x$ is a positive integer. the proof of the inequality requires some amount of calculus if $x$ is real. It can be proven without calculus if $x$ is rational. For the calculus based proof see the answer by Dr. MV. For rational $x$ see this answer http://math.stackexchange.com/a/1491888/72031 – Paramanand Singh Dec 11 '15 at 06:05
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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, $$ a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a) $$

user3258845
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