The discussion focuses on finding all positive integer solutions for the inequality $\dfrac{1}{4}<\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}<\dfrac{1}{2}$. Participants, including kaliprasad, engaged in solving the inequality by analyzing the bounds and deriving conditions for $x$. The conclusion identifies specific integer values that satisfy the inequality, demonstrating the application of rational expressions in number theory.
PREREQUISITESMathematicians, educators, and students interested in number theory and inequality solving techniques will benefit from this discussion.
kaliprasad said: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$