Find the positive integer solutions

[anemone]

Find all the positive integers $x$ such that $\dfrac{1}{4}<\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}<\dfrac{1}{2}$.

 

[kaliprasad]

Spoiler

we have
$\dfrac{1}{x}+ \dfrac{1}{x+2}$
=$ \dfrac{2x+2}{x^2+2x}$
$\gt \dfrac{2x+2}{x^2+2x+1}$
$\gt \dfrac{2(x+1)}{(x+1)^2}$
$\gt \dfrac{2}{x+1}$
hence
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\gt\dfrac{3}{x+2}$
further
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\lt \dfrac{3}{x+1}$
so to get the upper bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{4}\gt \dfrac{3}{x+2}$
so $10\lt x \lt 11$
so $x\le10\cdots(1)$
so to get the lower bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{2}\gt \dfrac{3}{x+2}$
so $4\lt x\lt 5$
so $x\ge5\cdots(2)$
from (1) and (2) $ 5 \le x \le 10$

 

[anemone]

Spoiler

we have
$\dfrac{1}{x}+ \dfrac{1}{x+2}$
=$ \dfrac{2x+2}{x^2+2x}$
$\gt \dfrac{2x+2}{x^2+2x+1}$
$\gt \dfrac{2(x+1)}{(x+1)^2}$
$\gt \dfrac{2}{x+1}$
hence
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\gt\dfrac{3}{x+2}$
further
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\lt \dfrac{3}{x+1}$
so to get the upper bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{4}\gt \dfrac{3}{x+2}$
so $10\lt x \lt 11$
so $x\le10\cdots(1)$
so to get the lower bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{2}\gt \dfrac{3}{x+2}$
so $4\lt x\lt 5$
so $x\ge5\cdots(2)$
from (1) and (2) $ 5 \le x \le 10$

Well done, kaliprasad! And thanks for participating!