Proving an Inequality by Induction: $n! < (n/2)^n$

Solution 1:

If $P(n)$ is the proposition you can show directly $P(6)$ is true.

Now, $P(k)$ is $k! < (k/2)^k$ which implies $k!(k+1) < (k/2)^k(k+1)$ and

$$ \begin{align} \left( \frac k2 \right)^k (k+1) &\leq \left( \frac{k+1}{2} \right)^{k+1} \ \Leftrightarrow \\k + 1 &\leq \frac{k+1}{2} \left( 1 + \frac 1k \right)^k \Leftrightarrow \\ 2 &\leq \left( 1 + \frac 1k \right)^k \end{align} $$

It is well known that $a_k = \left( 1 + \frac 1k \right)^k$ is a monotonically increasing sequence converging to $e$, greater than $2$ for all $k \geq 2$. Hence $P(k) \Rightarrow P(k+1)$ for all $k \geq 6$.

Therefore $P(n)$ for all $n \geq 6$.