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

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  • $\begingroup$ Did you mean to write $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$? The large exponent is really confusing. $\endgroup$ Jul 2, 2013 at 3:12
  • $\begingroup$ yes,Thank you,@CameronWilliams $\endgroup$
    – math110
    Jul 2, 2013 at 3:23
  • $\begingroup$ 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. $\endgroup$
    – vadim123
    Jul 2, 2013 at 5:12
  • $\begingroup$ No,let $x=4.4$ you can see:wolframalpha.com/input/… $\endgroup$
    – math110
    Jul 2, 2013 at 6:11
  • $\begingroup$ @math110, that is with k=1. With k=5 you have wolframalpha.com/input/… $\endgroup$
    – vadim123
    Jul 2, 2013 at 13:13

2 Answers 2

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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 $$f(x+1)+\frac{1}{f(x)}>f(x)+\frac{1}{f(x)}\ge 2$$

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  • $\begingroup$ yes,But for $x\ge e-1$? How to prove it? Thank you $\endgroup$
    – math110
    Jul 2, 2013 at 6:02
  • $\begingroup$ If I knew how (other than the original method), I would tell you. $\endgroup$
    – vadim123
    Jul 2, 2013 at 16:50
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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

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  • $\begingroup$ he has edit,@vadim123 $\endgroup$
    – math110
    Jul 2, 2013 at 4:56
  • $\begingroup$ oh,Now This methods also is wrong $\endgroup$
    – math110
    Jul 2, 2013 at 6:00
  • $\begingroup$ because $x>e$,then $f(x)>f(x+1)$,so $f(x+1)+\dfrac{1}{f(x)}$ can't prove than $2$ $\endgroup$
    – math110
    Jul 2, 2013 at 6:32
  • $\begingroup$ ya, there is a stupid mistake and I correct it now. thanks all who pointed out the msitake. $\endgroup$
    – chenbai
    Jul 2, 2013 at 7:35
  • $\begingroup$ Now this method is the same as in the OP; math110 asked for a different method. $\endgroup$
    – vadim123
    Jul 2, 2013 at 13:16

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