I tried using Induction, but I couldn't prove the inequality. Any proof would work.
Rewriting the question for clarity, here is its statement:
For any positive integer $n > 2$, prove that $n^{n/2} < n!$
I tried using Induction, but I couldn't prove the inequality. Any proof would work.
Rewriting the question for clarity, here is its statement:
For any positive integer $n > 2$, prove that $n^{n/2} < n!$
The key observation is that for all $2 \leq k \leq n-1$ it holds that $k\cdot(n-k)>n$.
Let's say $n$ is even. Then you can write $$ (n!)^2=(1\cdot n)(2\cdot (n-1))(3\cdot(n-3)... $$ We have $n/2$ terms in the product. The first product is equal to $n$. All others are greater than $n$.
For odd $n$ a similar argument works, but you need to isolate the middle element in the product.
From Stirling's Inequality,
$$ \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n+1}}<n! $$
Now using induction show that $\left(\frac{n}{e}\right)^n$ is greater than $n^{n/2}, \forall n>2$. The other terms on the LHS of the inequality are greater than 1 $\forall n>2$.