Show that among any seven distinct positive integers not greater than 126 one can find two of them, say x and y, satisfying the inequalities $1 < \dfrac y x \le 2.$
This is what I have done
Notice that $ 126/7 = 18$, therefore the maximum spread of all the numbers is $1, 19,37,55,73,91,109$. Therefore we are trying to solve the inequality $1<(x+18)/x<2$
therefore $x>18$ Hence whenever $x>18$ the inequality holds, however when $x<18$ the maximum range between the numbers is $18/7$ which is approximately equal to $3$.
Therefore we are solving the inequality $1<(x+3)/x<2$. Trivially $x>3$ and if the gap between numbers is greater than three than there is at least one pair of numbers with a small enough range to satisfy the initial inequality.
This is what i came up with, if I am in the right direction please suggest improvements if not please indicate where I have gone wrong thanks!