1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Is this question really belongs to sequences and series $\endgroup$ – John Dec 21 '16 at 17:05
  • $\begingroup$ @John: I don't see it. $\endgroup$ – robjohn Dec 21 '16 at 17:06
  • $\begingroup$ Actually It does. $\endgroup$ – Harsh Kumar Dec 21 '16 at 17:10
  • 3
    $\begingroup$ The question may come from a sequences and series situation, but as written, it doesn't seem to relate sequences and series. $\endgroup$ – Michael Burr Dec 21 '16 at 17:11
  • 1
    $\begingroup$ @HarshKumar Could you provide some reasoning behind why you believe it belongs there? $\endgroup$ – Kitter Catter Dec 21 '16 at 17:24
12
$\begingroup$

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 $1$.

$\endgroup$
5
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.