Showing if $n \ge 2c\log(c)$ then $n\ge c\log(n)$ Is this true that if $n \ge 2c\log(c)$ then $n\ge c\log(n)$, for any constant $c>0$?
Here $n$ is a positive integer.
 A: If $0\lt n\le1$, then for any $c\ge0$, we have
$$
n\ge c\log(n)\tag{1}
$$
Therefore, we can assume that $n\gt1$.
Since $\frac{n}{\log(n)}\ge e$ for all $n\gt1$, if $c\le e$, then automatically
$(1)$ true. Therefore, we can assume that $c\gt e$.
Suppose that $n\ge 2c\log(c)$. Since $n\ge 2e$, we have that $\frac{n}{\log(n)}$ is monotonically increasing. Therefore, we know that
$$
\begin{align}
\frac{n}{\log(n)}
&\ge\frac{2c\log(c)}{\log(c)+\log(2\log(c))}\\
&=c\frac2{1+\frac{\log(2\log(c))}{\log(c)}}\tag{2}
\end{align}
$$
Since $\frac{2\log(c)}c\le\frac2e\lt1$ for all $c\gt0$, we have that $2\log(c)\lt c$ and therefore,
$$
0\lt\log(2\log(c))\lt\log(c)\tag{3}
$$
$(2)$ and $(3)$ imply that
$$
\frac{n}{\log(n)}\ge c\tag{4}
$$
which implies $(1)$.
A: Suppose $n\ge 2c\ln c$ with $n,c>0$; and assume instead that $n<c\ln n$.
Then $\displaystyle\ln n>0$ and so $\displaystyle c>\frac{n}{\ln n},\;\;$ and $\displaystyle c\ln n>n\ge 2c\ln c \implies\ln n>2\ln c=\ln c^2\implies n>c^2.$
Therefore $\displaystyle c\ln n>n>c^2\implies\ln n>c>\frac{n}{\ln n}\implies(\ln n)^2>n\implies \ln n>\sqrt{n},\;$ and 
this gives a contradiction since $\ln x<\sqrt{x}$ for all $x>0$. ${\hspace .2 in}$  Therefore $n\ge c\ln n$.

(Notice that $\ln x<\sqrt{x}$ for $x>0$ since $f(x)=\sqrt{x}-\ln x$ has $\;f(4)=2(1-\ln2)>0$ as its minimum.)
