Show that $2 < e^{1/(n+1)} + e^{-1/n}$ I'm trying to show $2 < e^{1/(n+1)} + e^{-1/n}$. 
I can show that $ 2 < e^{1/n} + e^{-1/(n+1)}$ since
$$2 \leq 2\cosh\left(\frac{1}{n}\right) = e^{1/n} + e^{-1/n} < e^{1/n} + e^{-1/(n+1)}$$
but I'm still having trouble with the other inequality. I though using $\cosh$ again might help but I can't get anywhere. 
 A: Lemma. If $xy=1$ then $e^{-x}+e^{-y} \ge \frac2e$, with equality iff $x=y=1$.
Applying this lemma with $x=\frac{n}{n+1}$ and $y=\frac{n+1}{n}$ yields
$$ e^{1/(n+1)} + e^{-1/n}
= e(e^{\frac1{n+1}-1} + e^{-\frac1n-1})
= e(e^{-n/(n+1)} + e^{-(n+1)/n})
> 2 $$
as desired.
Proof of lemma.  [My previous proof of this lemma was wrong; it established the convexity of $e^{-e^t}$ on $(0,\infty)$ but then invoked it for $(-\infty,\infty)$.  This proof is more technical but, I believe, correct.]
If $x<0$ then $e^{-x}+e^{-y} > e^{-x} > 1 > \tfrac2e$.  From now on, assume $x>0$.
First, $e^{-x}+e^{-1/x}$ has a global minimum on $(0,\infty)$; here is a standard argument.  Let $b>0$ be such that $e^{-x}+e^{-1/x}>\frac12(1+\frac2e)$ whenever $x>b$; such $b$ exists because $e^{-x}+e^{-1/x}\to 1 > \frac12(1+\frac2e)$ as $x\to\infty$.  Similarly, let $a>0$ be such that $e^{-x}+e^{-1/x}>\frac12(1+\frac2e)$ whenever $0<x<a$.  By continuity and compactness, $e^{-x}+e^{-1/x}$ attains a minimum $M$ for $x\in[a,b]$.  Since $e^{-1}+e^{-1/1} = \frac2e < \frac12(1+\frac2e)$, we know that $1\in[a,b]$; thus also $M\le\frac2e<\frac12(1+\frac2e)$.  Since $e^{-x}+e^{-1/x} > \frac12(1+\frac2e)$ for all $x\notin[a,b]$, it follows that $M$ is a global minimum.
Equivalently, $e^{-x}+e^{-y}$ has a global minimum on the curve $xy=1$, $x>0$.  I will show that $e^{-x}+e^{-y}$ has only one critical point on this curve, namely $x=y=1$; it follows that this is the global minimum, as desired.
To find the critical points, we use Lagrange multipliers, which give the system
$$ \left\{\begin{aligned}
xy &= 1 \\
\lambda x &= e^{-y} \\
\lambda y &= e^{-x}
\end{aligned}\right. $$
Thus $\lambda^2 = \lambda^2 xy = e^{-x-y}$, and so $e^{-2y} = \lambda^2 x^2 = x^2 e^{-x-y}$, which since $x>0$ yields
$$ x = e^{(x-y)/2} $$
Since $xy=1$, we also have
$$ y = e^{(y-x)/2} $$
If $x\ne y$, then by the inequality of the geometric and logarithmic means,
$$ 1 = \sqrt{xy} < \frac{x-y}{\log x - \log y}
= \frac{x-y}{\frac{x-y}2 - \frac{y-x}2} = 1 $$
which is absurd.  Thus any critical point must satisfy $x=y$, as claimed.
A: well, since 
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+2k}} -\frac1{(4+2k+1)!n^{4+2k+1}}\right)>0 $$
and
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+k}} -\frac1{(4+2k+1)!n^{4+2k}}\right) +\frac1{(4+2k+2)!n^{4+2k+2}} >0 $$
then, 
$$\sum_{k\geq4}\frac{(-1)^k}{k!n^k}>0$$
we have (or the lemma below)
$$e^{-\frac1n}=1-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\sum_{k\geq4}\frac{(-1)^k}{k!n^k}\gt 1-\frac1n+\frac1{2n^2}-\frac1{6n^3}$$
it is obvious that
$$e^{\frac1{1+n}}\gt 1+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}$$
if $n>2$, it is easy to check that
$$e^{-\frac1n}+e^{\frac1{1+n}}\gt 2-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}\\ >2$$
In fact 
$$n^3(n+1)^3\left(-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{1+n}+\frac1{2{(1+n)^2}}+\frac1{6{(1+n)^3}}+ \frac1{24{(1+n)^4}}\right)=\frac{n^3}{24(n+1)}-\frac16=\frac{n(n^2-4)-4}{24(n+1)}\gt0$$

edit: 

Lemma 

$x\ne0$, $k$ is a odd number, then

$$e^x\gt 1+x+\frac{x^2}{2!}+\dotsb+ \frac{x^k}{k!}$$

Proof  $\qquad$ according to Taylor formula, there exists $\theta(0\lt\theta\lt1)$ such that

$$e^x= 1+x+\frac{x^2}{2!}+\dotsb+ \frac{x^k}{k!}+\frac{e^{\theta x}}{(k+1)!}x^{k+1}$$



