Prove that $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2$ for $x > 0$ 
Let $x>0$. Show that
$$(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2.$$

Do you have any nice method?
My idea $F(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$
then we hvae $F'(x)=\cdots$ But it's ugly.
can you have nice methods? Thank you
by this I have see this same problem
let $0<x<1$ we have $$x+\dfrac{1}{x^x}<2$$
this problem have nice methods:
becasue we have
$$\dfrac{1}{x^x}=\left(\dfrac{1}{x}\right)^x\cdot 1^{1-x}<x\cdot\dfrac{1}{x}+(1-x)\cdot 1=2-x$$
 A: This is a partial solution for $0<x<e-1$.
First note that for positive $y$, $y+\frac{1}{y}\ge 2$.
The function $f(x)=x^{1/x}$ increases on $(0,e)$, and decreases on $(e,\infty)$; this can be checked by considering $f'(x)$.
Hence if $x+1\le e$, then $$f(x+1)+\frac{1}{f(x)}>f(x)+\frac{1}{f(x)}\ge 2$$
A: First to prove $f(x)=x^{\frac{1}{x}}$ have a max $f(e)$.
Edit:2nd version
when $x<e-1, f(x+1)>f(x)$
when $x>e,f(x)>f(x+1)$
for $f(x+1)>f(x)$, it is trivial .
when $e-1\le x\le e$, : 
$f(x) $ is mono increasing, so $f(x)^{-1}_{min}=f(e)$, $f(x+1)$ is mono decreasing, so $ f(x+1)_{min}=f(e+1)$, thus,$f(x+1)+f(x)^{-1} > (e+1)^{\frac{1}{e+1}}$$+\dfrac{1}{e^{\frac{1}{e}}}=2.11 >2$
for $x>e$, let $g(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$, I will prove g(x) is mono decreasing,so $g(x)_{min}=g(+\infty)=2$
$g'(x)=x^{-2 - \frac{1}{x}} (-1 + Ln(x)) - (1 + x)^{-2 + \frac{1}{1 + x}} (-1 + Ln(1 + x))$, now to prove:
$x^{-2 - \frac{1}{x}}<(1 + x)^{-2 + \frac{1}{1 + x}} \iff \dfrac{x^2x^{\frac{1}{x}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \dfrac{x^2(x+1)^{\frac{1}{x+1}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \left(\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1} \right)^2>1 \iff\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1}>1 \iff \dfrac{1}{x+1}Ln(x+1)>Ln(x+1)-Ln(x) \iff (x+1)Ln(x)>xLn(x+1) \iff x^{\frac{1}{x}} > (x+1)^{\frac{1}{x+1}} $
so $g'(x)<0$ when $x>e$. Done
A: Remark: Bernoulli inequality and AM-GM are enough.

Letting $x = \frac{1}{y}$, it suffices to prove that, for all $y > 0$,
$$\left(1 + \frac{1}{y}\right)^{\frac{y}{1+y}} + y^y > 2.$$
If $y \ge 3$, clearly the inequality is true.
If $0 < y < 3$, we have
\begin{align*}
 \left(1 + \frac{1}{y}\right)^{\frac{y}{1+y}} + y^y - 2
 &= \frac{(1+y)^{\frac{y}{1+y} + 1}}{(1+y)\cdot (y^y)^{\frac{1}{1+y}}} + y^y - 2\\[5pt]
 &\ge \frac{1 + y \left(\frac{y}{1+y} + 1\right)}{(1+y)\cdot \left(1 + (y^y - 1)\cdot \frac{1}{1+y}\right)} + y^y - 2 \tag{1}\\[5pt]
 &= \frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} + y^y - 2\\[5pt]
 &= \frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} + (y^y + y) - y - 2\\[5pt]
 &\ge 2\sqrt{\frac{2y^2 + 2y + 1}{(1+y)(y^y + y)} \cdot (y^y + y)} - y - 2 \tag{2}\\[5pt]
 &= 2\sqrt{\frac{2y^2 + 2y + 1}{1 + y}} - y - 2\\[5pt]
 &> 0
\end{align*}
where in (1) we have used Bernoulli inequality to obtain
$$(1+y)^{\frac{y}{1+y} + 1} \ge 1 + y \left(\frac{y}{1+y} + 1\right)$$
and
$$(y^y)^{\frac{1}{1+y}} \le 1 + (y^y - 1)\cdot \frac{1}{1+y};$$
and in (2) we have used AM-GM,
and the last inequality follows from
$$\left(2\sqrt{\frac{2y^2 + 2y + 1}{1 + y}}\right)^2 - (y + 2)^2 = \frac{y^2 (3 - y)}{1 + y} > 0.$$
We are done.
