297. Nice inequalities with concrete numbers.

by Virgil Nicula, Jul 14, 2011, 1:27 AM

PP1 (Cristinel Mortici). Prove that $1.01^{1000}>1000$ .

Proof. $1.01^{1000}=$ $\left[\left(1+\frac {1}{100}\right)^{100}\right]^{10}\ \stackrel{(\mathrm{Bernoulli})}{\ge}\ \left(1+100\cdot\frac {1}{100}\right)^{10}=$ $2^{10}=1024>1000$ .

Extension. Let $\{p,m\}\subset\mathbb N$ so that $m\ge p+2$ . Prove that $1.000\ldots 001^{10^m}>1000^{10^{m-p-2}}$ ,

where the base from the left side has $p$ zeroes after point, i.e. $1.000\ldots 001=1+\frac {1}{10^{p+1}}$ .

Proof. $1.000\ldots 001^{10^m}=$ $\left[\left(1+\frac {1}{10^{p+1}}\right)^{10^{p+1}}\right]^{10^{m-p-1}}\ \stackrel{(\mathrm{Bernoulli})}{\ge}\ \left(1+10^{p+1}\cdot\frac {1}{10^{p+1}}\right)^{10^{m-p-1}}=$

$2^{10^{m-p-1}}=\left(2^{10}\right)^{10^{m-p-2}}=1024^{10^{m-p-2}}>1000^{10^{m-p-2}}$ .

Particular case. $p=1$ and $m=3\implies 1.01^{1000}>1000$ or more generally $ \left(1+\frac {1}{10^p}\right)^{10^{p+1}}>1000$ for any $p\in\mathbb N$ .

PP2. Prove that for any $n\in\mathbb N^*\ ,\ n\ge 3$ exists the inequality $\left(n!\right)^2\ >\ n^n$ .

Proof 1. Observe that for any $k\in \overline{1,n}$ exists the inequality $(k-1)(k-n)\le 0\iff$ $k(n+1-k)\ge n$ . Thus, $\prod_{k=1}^n k(n+1-k)> n^n\iff (n!)^2> n^n$ for any $n\ge 3$ .

Proof 2. Denote $a_n=\frac {(n!)^2}{n^n}$ . Thus, $\frac {a_{n+1}}{a_n}=\frac {n+1}{\left(1+\frac 1n\right)^n}>\frac {n+1}{e}>1$ for $n\ge 2$ . In conclusion, for any $n\ge 2\ ,\ a_{n+1}>a_n\ge a_2=1\implies$ $(n!)^2> n^n$ for any $n\ge 3$ .
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:52 AM

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