Problem - 2218
Show that if $k \ge 4$, then $lcm(1; 3;\cdots; 2k- 3; 2k- 1) > (2k + 1)^2$ where $lcm$ stands for least common multiple.
It is easy to show that $2k-1, 2k-3, 2k-5$ are coprime. Therefore,
$$
\begin{array}{ll}
&lcm(1; 3; \cdots ; 2k-3; 2k-1)\\
&\ge lcm(2k-1; 2k-3; 2k-5)\\
& = (2k-1)(2k-3)(2k-5)\\
& \ge 7(2k-3)(2k-5)\\
& > 6(2k-3)(2k-5) \\
& = (4k-6)(6k-6) \\
& > (2k+1)(2k+1)\\
& = (2k+1)^2
\end{array}
$$