<s>[INST] Trying to prove $\log\left(n!\right)$ is greater than or equal to $\left(\frac{n}{2}\log\left(n\right)\right)$ I'm trying to prove that $\log\left(n!\right)$  is greater than $\left(\frac{n}{2}\log\left(n\right)\right)$ and I'm kinda stuck.
I could only prove $\log\left(n!\right)\ge\ \left(\frac{n}{2}\log\left(\frac{n}{2}\right)\right)$ and following is my prove:
\begin{align}
\log(n!) &=\log(n\cdot(n-1)\cdot(n-2)\cdots(1))\\ 
&=\log(n)+\log(n-1)+\log(n-1)+\cdots+\log(1) \\
&\geq \log(n)+\cdots+\log(\frac{n}{2}) \\
&\geq \log(\frac{n}{2})+\log(\frac{n}{2})+\log(\frac{n}{2})+\cdots+\log(\frac{n}{2})\\
&= \frac{n}{2}\log(\frac{n}{2})
\end{align}
But this is not helping me either because $\frac{n}{2}\log(n) > \frac{n}{2}\log(\frac{n}{2})$ 
and $\frac{n}{2}\log(\frac{n}{2}) = \frac{n}{2}\log(n) - \frac{n}{2}\log(2)$
What am I missing?
 [/INST] You're almost there -- just don't discard the second half of the sum (and deal with the "good" term $\log n$ separately):$$\begin{align}
\log(n!) &= \sum_{k=1}^n \log k = \log n + \sum_{k=\frac{n}{2}+1}^{n-1} \log k + \sum_{k=2}^{\frac{n}{2}}\log k \\
&\geq \log n + \left(\frac{n}{2}-1\right)\log \frac{n}{2} + \left(\frac{n}{2}-1\right) \log 2 \\
&= \log n + \left(\frac{n}{2}-1\right)\log n - \left(\frac{n}{2}-1\right)\log 2 + \left(\frac{n}{2}-1\right) \log 2 \\
&= \frac{n}{2}\log n
\end{align}$$
</s>