Upper Bound of Logarithm For $1\leq x < \infty$, we know $\ln x$ can be bounded as following:
 $\ln x \leq \frac{x-1}{\sqrt{x}}$.
Then what is the upper bound of $\ln x$ for following condition?   
$2\leq x <\infty$ 

Given $a>0$, if $a \le u < \infty$ then also $1 \le u/a < \infty$, and you can apply your inequality taking $x=u/a$ to get
$$\ln(u/a)\le \frac{u/a-1}{\sqrt{u/a}}.$$
Then cleaning it up you have $$\ln(u)\le \ln(a)+\frac{u-a}{\sqrt{au}}.$$
This is a bit less appealing than the $a=1$ case wherein the logarithm doesn't appear in the bounding function, but actually it only appears in a constant.
