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If $x$ is a positive proper fraction. Prove that $$(1+x)^{1-x}(1-x)^{1+x}<1$$

Actually this question has two parts I can't do the $1^{st}$ part but the $2^{nd}$ part was quite easy with respect to the $1^{st}$ one. The $2^{nd}$ was to show that $$a^bb^a<(\frac{a+b}{2})^{a+b} $$ I observe that if $1^{st}$ is true then if I will put $$x=\frac{a-b}{a+b}$$ the equation of part $1^{st}$ will take the form $$a^bb^a<(\frac{a+b}{2})^{a+b} $$ and hence proved but I am unable to prove that $(1+x)^{1-x}(1-x)^{1+x}<1$

robjohn
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Harsh Kumar
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    Is this question really belongs to sequences and series – John Dec 21 '16 at 17:05
  • @John: I don't see it. – robjohn Dec 21 '16 at 17:06
  • Actually It does. – Harsh Kumar Dec 21 '16 at 17:10
  • This is a question of sequences and series – Harsh Kumar Dec 21 '16 at 17:11
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    The question may come from a sequences and series situation, but as written, it doesn't seem to relate sequences and series. – Michael Burr Dec 21 '16 at 17:11
  • Actually I also couldn't understand this, but it belongs to sequences and series – Harsh Kumar Dec 21 '16 at 17:13
  • Apply logarithm – lab bhattacharjee Dec 21 '16 at 17:23
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    @HarshKumar Could you provide some reasoning behind why you believe it belongs there? – Kitter Catter Dec 21 '16 at 17:24
  • I tried to use logarithm series in this and it worked. – Harsh Kumar Dec 21 '16 at 17:26
  • @HarshKumar: there is nothing about sequences or series in this question. If you were to show that you used series, then perhaps it might merit the [sequences-and-series] tag, but as it stands, it no more should have that tag than the [logarithms] tag because an answer used logarithms. – robjohn Dec 22 '16 at 09:43
  • @HarshKumar: the description of the [sequences-and-series] tag says "For questions about recurrence relations, convergence tests, and identifying sequences." The question has nothing to do with this. Since the question has not been modified to include anything relating it to this tag, I am removing the tag. If the question is amended to include some relation to [sequences-and-series], the tag might apply. – robjohn Dec 23 '16 at 13:50

4 Answers4

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Sketch: $$ (1+x)^{1-x}(1-x)^{1+x}=(1+x)^{1-x}(1-x)^{1-x}(1-x)^{2x}=(1-x^2)^{1-x}(1-x)^{2x} $$ Since both $1-x^2<1$ and $1-x<1$ (and the exponents are positive), this is a product of terms less than $1$.

Michael Burr
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I would use here the inequality $(1 + x) \leqslant e^x$. This gives us \begin{split} (1+x)^{1-x}(1-x)^{1+x} &\leqslant (e^x)^{1-x}(e^{-x})^{1+x} \\ &= e^{x-x^2}e^{-x-x^2} \\ &= e^{-2x^2} \\ &= \left(\frac{1}{e}\right)^{2x^2} \\ &< 1,\quad x>0 \end{split}

Vladimir Nechaev
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Since $\log$ is concave, $$ \begin{align} \frac{1-x}2\log(1+x)+\frac{1+x}2\log(1-x) &\le\log\left(\frac{1-x}2(1+x)+\frac{1+x}2(1-x)\right)\\ &=\log\left(1-x^2\right) \end{align} $$ Therefore, $$ \color{#090}{(1+x)^{1-x}(1-x)^{1+x}}\le\color{#C00}{\left(1-x^2\right)^2} $$

enter image description here

robjohn
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We just need prove:$$(1-x)\ln{(1+x)}+(1+x)\ln{(1-x)}<0$$ $$\Longleftrightarrow \frac{\ln{(1+x)}}{1+x}+\frac{\ln{(1-x)}}{1-x}<0$$ So,you need a function:$$f(x)=\frac{\ln{(1+x)}}{1+x}$$ I hope its useful for you

Zuo
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