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I asked this question yesterday and have been working on it. I have to prove that $2015^{2013}<2014^{2014}<2013^{2015}$.

I set $x=2014$, so now I have $$(x+1)^{x-1}<x^x<(x-1)^{x+1}.$$

Since $x^x=\exp(x\log x)$ I have to show that $$(x-1)\log{(x+1)}<x\log x<(x+1)\log{(x-1)}.$$

It's clear to me that $(x-1)< x < (x+1)$, but I don't know how to show that these values are increasing. I've tried taking the derivative values but still not making the connection. What am I missing?

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  • $\begingroup$ what is the difference between the variables $X$ and $x? $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '14 at 15:39
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    $\begingroup$ The function $\dfrac{x}{\ln x}$ is strictly increasing for all $x>e$. $\endgroup$ – user170039 Oct 21 '14 at 15:55
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Let's rewrite the inequality a bit as follows $$x\log(x+1)-x\log x< \log(x+1)\tag{1}$$ Let $h=\frac1x$, the LHS is $$\frac{\log(1+h)-\log1}{h}=\frac1{1+c}<1$$ for some $c\in (0,h)$. The inequality (1) follows for $x$ large enough ($>2$).

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Since $(1+\frac{1}{n})^n < e < (1+\frac{1}{n})^{n+1}$, we have $n\ln(1 + \frac{1}{n}) < 1 < (n+1)\ln(1 + \frac{1}{n})$, so $\frac{1}{n+1}<\ln(1+\frac{1}{n})<\frac{1}{n}$, or $\frac{1}{n+1}<\ln (n+1) - \ln n<\frac{1}{n}$. So

$$2013(\ln2015 - \ln2014) < \frac{2013}{2014} < \ln2014\implies $$$$2013\ln2015<2014\ln2014\implies 2015^{2013} < 2014^{2014}$$ and

$$2014(\ln2014 - \ln2013) < \frac{2014}{2013} < \ln2013\implies$$ $$2014\ln2014<2015\ln2013\implies 2014^{2014} < 2013^{2015}$$

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Hint: Note that $f(x+1)-f(x)=f'(\xi)$ for some $\xi$ with $x<\xi<x+1$ by the Mean Value Therorem.

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For example: for big $\;x>0\;$

$$(x+1)^{x-1}<x^x\iff \left(1+\frac1x\right)^x<x+1$$

and the rightmost inequality follows at once as the left side converges to $\;e\;$ whereas the right one diverges to infinity.

The other inequality is similar to this one.

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Define, for $x\in [-1,1]$, $f(x)= (2014-x)\log(2014+x)$. Then $$f'(x) = -\log(2014+x) + \dfrac{2014-x}{2014+x}$$

Now, $f''(x)= -\dfrac{1}{2014+x}- \dfrac{4028}{(2014+x)^2}$ and $f''(x)=0 \Longleftrightarrow x=-6042$. This means that for $x\in [-1,1]$, $f'(x)$ has no change in its sign. But $f'(0)=-\log(2014)+1<0$, and therefore $$f'(x)<0,\ \forall x\in[-1,1]$$

As a consequence, we have $f(x)$ is decreasing in $[-1,1]$ and finally $$f(1) < f(0) < f(-1)$$

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