# How to prove this inequality $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2$

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 $$\dfrac{1}{x^x}=\left(\dfrac{1}{x}\right)^x\cdot 1^{1-x}<x\cdot\dfrac{1}{x}+(1-x)\cdot 1=2-x$$

• Did you mean to write $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$? The large exponent is really confusing. – Cameron Williams Jul 2 '13 at 3:12
• yes,Thank you,@CameronWilliams – math110 Jul 2 '13 at 3:23
• This is a very delicate problem. If you replace $x+1$ by $x+k$, for any $k>4.047$, the inequality fails to hold. For example, if $k=5$, take $x=4.4$ and the sum is less than 2. – vadim123 Jul 2 '13 at 5:12
• No,let $x=4.4$ you can see:wolframalpha.com/input/… – math110 Jul 2 '13 at 6:11
• @math110, that is with k=1. With k=5 you have wolframalpha.com/input/… – vadim123 Jul 2 '13 at 13:13

First to prove $f(x)=x^{\frac{1}{x}}$ have a max $f(e)$.
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}}$$+\dfrac{1}{e^{\frac{1}{e}}}=2.11 >2 for x>e, let g(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}, I will prove g(x) is mono decreasing,so g(x)_{min}=g(+\infty)=2 g'(x)=x^{-2 - \frac{1}{x}} (-1 + Ln(x)) - (1 + x)^{-2 + \frac{1}{1 + x}} (-1 + Ln(1 + x)), now to prove: x^{-2 - \frac{1}{x}}<(1 + x)^{-2 + \frac{1}{1 + x}} \iff \dfrac{x^2x^{\frac{1}{x}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \dfrac{x^2(x+1)^{\frac{1}{x+1}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \left(\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1} \right)^2>1 \iff\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1}>1 \iff \dfrac{1}{x+1}Ln(x+1)>Ln(x+1)-Ln(x) \iff (x+1)Ln(x)>xLn(x+1) \iff x^{\frac{1}{x}} > (x+1)^{\frac{1}{x+1}} so g'(x)<0 when x>e. Done • he has edit,@vadim123 – math110 Jul 2 '13 at 4:56 • oh,Now This methods also is wrong – math110 Jul 2 '13 at 6:00 • because x>e,then f(x)>f(x+1),so f(x+1)+\dfrac{1}{f(x)} can't prove than 2 – math110 Jul 2 '13 at 6:32 • ya, there is a stupid mistake and I correct it now. thanks all who pointed out the msitake. – chenbai Jul 2 '13 at 7:35 • Now this method is the same as in the OP; math110 asked for a different method. – vadim123 Jul 2 '13 at 13:16 This is a partial solution for 0<x<e-1. First note that for positive y, y+\frac{1}{y}\ge 2. The function f(x)=x^{1/x} increases on (0,e), and decreases on (e,\infty); this can be checked by considering f'(x). Hence if x+1\le e, then$$f(x+1)+\frac{1}{f(x)}>f(x)+\frac{1}{f(x)}\ge 2$$• yes,But for$x\ge e-1\$? How to prove it? Thank you – math110 Jul 2 '13 at 6:02