I just began with combinatorics and so far, I've done nothing more than learn about the definition of the factorial and do a few questions based on it, one of which was this :
Prove that $\forall n\in\Bbb Z^+, (n!)^2\leq n^n(n!)<(2n)!$
I was wondering how and why one comes up with seemingly weird inequalities like these.
Pardon me if my question sounds absurd.
Thank You!
Identities like these are usually discovered by playing with strings of numbers. That's also a good way to discover the intuition behind the identity. For example take $n = 6$. The three expressions are:
$A = \left(6!\right)^2 = \left(1\cdot2\cdot3\cdot4\cdot5\cdot6\right)^2$
$B = 6^6 \left(6!\right) = \left(6\cdot6\cdot6\cdot6\cdot6\cdot6\right)\left(1\cdot2\cdot3\cdot4\cdot5\cdot6\right)$
$C = \left(2\cdot6\right)! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12$
We can go from $A$ to $B$ by replacing one of the $1$s with $6$, and one of the $2$s with $6$, ... and one of the $6$s with $6$. Each of these makes the number bigger, so $A \leq B$.
Similarly, we can go from $B$ to $C$ by replacing one of the $6$s in the first group with $7$, and one of the $6$s in the first group with $8$, ... and the last one of the $6$s in the first group with $12$. Each of these makes the number bigger, so $B \lt C$.
THIS IS NOT A PROOF but it shows WHY the identity is true.