{n\sqrt6 } > 1/n and {n\sqrt6 }> / 1/ (n-1/(5n)) , for odd n

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

{n\sqrt6 } > 1/n and {n\sqrt6 }> / 1/ (n-1/(5n)) , for odd n

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2014 Belarus TST 8.2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30627 posts

#1 h Dec 29, 2020, 8:19 AM

Y by

Prove that for all even positive integers $n$ the following inequality holds
a) $\{n\sqrt6\} > \frac{1}{n}$
b) $\{n\sqrt6\}> \frac{1}{n-1/(5n)}$

(I. Voronovich)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4867 posts

#2 h Dec 29, 2020, 9:06 AM

Y by

$n=2k$
Let $[2k\sqrt{6}]=m$
Case 1: $m=2t$
Then $24k^2-4t^2=4(6k^2-t^2) \geq 8$
Case 2: $m=2t+1$
$24k^2-(2t+1)^2=c$
$24k^2-4t^2-4t=c+1 \to 8|c+1 \to c \geq 7$
If $c=7$ then $6k^2=t^2+t+2$ has no solutions, so $c \geq 15$

So $24k^2-m^2 \geq 8$
$\{2k\sqrt6\}=2k\sqrt{6}-[2k\sqrt{6}]=\frac{24k^2-m^2}{2k\sqrt{6}+m} >\frac{4}{2k\sqrt{6}}=\frac{4}{n\sqrt{6}}$

b ) $\frac{4}{n\sqrt{6}}> \frac{1}{n-\frac{1}{5n}}$
$4(5n^2-1)>5\sqrt{6}n^2$
$5(4-\sqrt{6})n^2>4$ - true for all $n$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

507 posts

Jjesus

#3 h May 15, 2021, 4:08 PM • 3 Y

Y by Mango247, Mango247, Mango247

RagvaloD wrote:

$\frac{24k^2-m^2}{2k\sqrt{6}+m} >\frac{4}{2k\sqrt{6}}$

Why?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

507 posts

Jjesus

#4 h Jun 29, 2021, 9:53 PM

Y by

any ideas?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9022 posts

#5 h Jun 30, 2021, 8:11 AM

Y by

Well, as shown in the argument the numerator is at least $8$ , so the fraction is at least $\frac{8}{2k\sqrt{6}+m}>\frac{8}{4k\sqrt{6}}=RHS$ since $m<2k\sqrt{6}$ by definition.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

579 posts

#6 h Jun 30, 2021, 5:28 PM

Y by

Claim 1: The following terms are never squares:
$24k^2, 24k^2-1,24k^2-2,24k^2-3,24k^2-4,24k^2-5,\ldots 24k^2-7$ Proof: The only quadratic residues mod 24 are 0,1,4,9,12,16, which deal with $24k^2-1,\ldots 24k^2-7$ . Then, note that $24k^2$ can never be a square because $v_3(24k^2)\equiv 1\pmod{2}$ .

Claim 2:
For positive even integers $n$ ,
$\sqrt{6n^2}- \sqrt{6n^2-8}>\frac{1}{n-\frac{1}{5n}}$ Proof:
Note that,
$\frac{1}{n-\frac{1}{5n}}^2 + 2\cdot \frac{1}{n-\frac{1}{5n}} \cdot \sqrt{6n^2-8}+6n^2-8 < 1+ 2\cdot \frac{1}{\sqrt{n^2-\frac{8}{6}} } \sqrt{6n^2-8} +6n^2-8$ $=1+2\sqrt{6}+6n^2-8 < 6n^2$ $\square$

Applying Claim 1, for even $n$ ,
$\lfloor \sqrt{6}n \rfloor \leq \sqrt{6n^2-8}$ Then, applying Claim 2,
$\{n\sqrt{6}\}=n\sqrt{6}-\lfloor \sqrt{6}n\rfloor \geq \sqrt{6n^2}- \sqrt{6n^2-8}>\frac{1}{n-\frac{1}{5n}}$ $\blacksquare$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a