Prove if $f(x) = 4x^2 +9$, $f(x) = \Omega (x \log x)$

Question

How do I prove this using inequalities? I know that since $4x^2 +9= \Omega (x^2)$, it follows that $4x^2 +9$ is $\Omega (x \log x)$ since Big Omega is a lower bound. But how do I prove it using inequalities and choose a $n_0$ and c? I don't really know how to get from $4x^2 +9$ to $\Omega (x^2)$, or $\Omega (x \log x)$

Answer 1

The relation $4x^2 +9= \Omega (x^2)$ follows simply considering $c=4$ and then $\forall x \geq 0$ it holds $4x^2 +9 \geq cx^2$. Then, to show that $x^2 = \Omega(x\log x)$ it suffices to prove that $x = \Omega(\log x)$ which easily follows since $\forall x > 0$ $x > \log x$.