Show that $(1+x)^{1-x}(1-x)^{1+x}<1$

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 $(1+x)^{1-x}(1-x)^{1+x}<1$

• 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
• 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
• @HarshKumar Could you provide some reasoning behind why you believe it belongs there? – Kitter Catter Dec 21 '16 at 17:24

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 $1$.
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}
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, $$\color{#090}{(1+x)^{1-x}(1-x)^{1+x}}\le\color{#C00}{\left(1-x^2\right)^2}$$
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