how to prove that $n\log n>n$ and $\sqrt{n}<\log(n)$?

Question

How do I prove that for some $n_0$ and for all $n>n_0$:

1. $n\log n>n$

2. $\sqrt{n}<\log(n)$?

Answer 1

1) divide by $n$ and you get $\ln(n)>1$ that is true for $n\geq 3=n_0$ (the minimum integer greater than $e$).

2) is false. You can see that the opposite inequality ($>$) is true for $n\geq 1=n_0$: taking the continuous counterpart $f(x)=\sqrt{x}$ and $\ln(x)$ for $x\geq1$ and show that $f'(x)>g'(x)$ for $x>1$ and $f(1)>g(1)$. Conclude applying the fundamental theorem of calculus and monotony of the integral with respect to the integrand.