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Prove $$0.9999^{\!101}<0.99<0.9999^{\!100}$$ I think its original idea is $$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$ For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.

Furthermore $0.99^{\!-99}\times0.99^{\!100}=0.99<1.01^{\!100}\times0.99^{\!100}=0.9999^{\!100}$.

For $x=-100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,1.01^{\!101}<0.99^{\!-100}$.

Furthermore $1.01^{\!101}\times0.99^{\!101}=0.9999^{\!101}<0.99^{\!-100}\times0.99^{\!101}=0.99$.

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  • $\begingroup$ It's not too hard to prove this for the two exponential parts by taking the log of each side. I'm curious if a similar approach can be used for the middle. $\endgroup$
    – Max
    May 28 '19 at 7:06
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$$0.99<0.9999^{100}$$ it's $$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli: $$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$ $$0.9999^{101}<0.99$$ it's $$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or $$\left(1-\frac{1}{100^2}\right)^{100}\left(1+\frac{1}{100}\right)<1$$ or $$\left(1-\frac{1}{100^2}\right)^{100}<1-\frac{1}{101},$$ which is true because $$\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}<1-\frac{1}{101}.$$ I used the following lemma.

Let $0<x<1$ and $n>2.$ Prove that: $$(1-x)^n<1-nx+\frac{n(n-1)}{2}x^2.$$

Proof.

We need to prove that $f(x)>0,$ where $$f(x)=1-nx+\frac{n(n-1)}{2}x^2-(1-x)^n.$$ Indeed, $$f'(x)=-n+n(n-1)x+n(1-x)^{n-1}$$ and $$f''(x)=n(n-1)-n(n-1)(1-x)^{n-2}=n(n-1)\left(1-(1-x)^{n-2}\right)>0.$$ Thus, $$f'(x)>f'(0)=0,$$ $$f(x)>f(0)=0$$ and the lemma is proven.

Now, take $x=\frac{1}{100^2}$ and $n=100.$

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  • $\begingroup$ What about $0.9999^{101}<0.99?$ $\endgroup$ May 28 '19 at 4:30
  • $\begingroup$ @J. W. Tanner I added something. See now. $\endgroup$ May 28 '19 at 4:33
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    $\begingroup$ Thank you very much $\endgroup$ May 28 '19 at 4:34
  • $\begingroup$ How to show that $\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}$. $\endgroup$ May 28 '19 at 4:35
  • $\begingroup$ @zongxiang yi I added something. See now. $\endgroup$ May 28 '19 at 4:46
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Here is an outline of a proof that $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1,$

using some of your tags such as derivatives and logarithms.

Let $f(x)=\dfrac{(1+x)^{\frac1x}}{(1-x)^{(1-\frac1x)}}=\dfrac{(1-x^2)^{\frac1x}}{1-x}.$

Note that $f(0)$ is not defined, but $\lim_{x\to0}f(x)=1$.

$f'(x)=-(1-x^2)^{\left(\frac1x-1\right)}\left[1+\dfrac{1+x}{x^2}\ln(1-x^2)\right].$

For $0<x<1$, $\ln(1-x^2)<-x^2<-\dfrac{x^2}{1+x}$, so $f'(x)>0.$

For $-1<x<0, \ln(1-x^2)>-\dfrac{x^2}{1+x},$ so $f'(x)<0.$

It follows that $f(x)>1$ when $0<|x|<1,$

and thus $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1.$

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