Prove: $\frac{n}{3}\log n - \frac{n}{2}\log \sqrt{n}$ is $\Omega(n\ \log n) $.

I'm trying to use the Big-Omega definition, what I reached so far is:


$f(n)=\frac{n}{3}\log n - \frac{n}{2}\log \sqrt{n}$

$g(n) = (n\log n)$

Then, $f(n)$ is in $\Omega(g(n))$ if there is $c>0$ and $n_0 > 0$ such that $f(n) \geq c \times g(n)$ for some $n\geq n_0$

How do I continue from here?



$$\frac{n}{3}\log\ n\ - \frac{n}{2}\log \sqrt{n} = \frac{n}{3}\log\ n\ - \frac{n}{4}\log {n} = \frac1{12}\, \left(n\log\ n\right)$$

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