Solving $2\log(n!) - n\log n \le 2n\log n$

Question

I need to prove that: $$2log(n!) - nlogn \le 2nlogn$$

Can I say that: $$2log(n!) - nlogn \le 2log(n!) = 2\cdot \Sigma^n_{i=1} log(i) \color{red}{\le} 2\cdot\Sigma^n_{i=1} log(n)$$ $$\Sigma^n_{i=1} log(n) \le 2n\cdot logn$$

Is the $\color{red}{red}$ inequality holds? I really can't tell if this is right. Would love for some help. Thank you.

Answer 1

For $n\ge 1,$ we have

$$1\le n!\le n^n$$ thus $$(n!)^2\le (n^n)^2\le (n^n)^3$$

taking logarithm, we get

$$2\ln(n!)\le 3n\ln(n)$$

and $$2\ln(n!)-n\ln(n)\le 2n\ln(n)$$ done.