Calculus Proof of inequality: $2015^{2013}<2014^{2014}<2013^{2015}$ 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?
 A: 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 ($>2$).
A: 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
$$2014(\ln2014 - \ln2013) < \frac{2014}{2013} < \ln2013\implies$$
$$2014\ln2014<2015\ln2013\implies 2014^{2014} < 2013^{2015}$$
A: Hint: Note that $f(x+1)-f(x)=f'(\xi)$ for some $\xi$ with $x<\xi<x+1$ by the Mean Value Therorem.
A: 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.
A: 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 $$f(1) < f(0) < f(-1)$$ 
