<s>[INST] 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?   
$2\leq x <\infty$ 
 [/INST] To get your first upper bound we may start with the expansion :
$$\ln(1+x)=x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 + O(x^5)$$
with the upper bound :
$$\frac x{\sqrt{1+x}}=x - \frac{x^2}2 + \frac{3x^3}8 - \frac{5x^4}{16} + O(x^5)$$
To stay in the same spirit you may use the expansion :
$$\ln\left(1+\frac x2\right)=\frac x2 - \frac{x^2}8 + \frac{x^3}{24} - \frac{x^4}{64} + O(x^5)$$
and more generally (for $a$ positive) :
$$\ln\left(1+\frac xa\right)=\frac xa - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + O(x^5)$$
we could use the same formula for the upper bound (so that the initial proof remains valid) :
$$\frac {\frac xa}{\sqrt{1+\frac xa}}= \frac xa - \frac{x^2}{2a^2} + O(x^3)$$
Let's use $\ln\left(1+\frac xa\right)=\ln(a+x)-\ln(a)$ and $z:=a+x$ to get :
$$\ln(z)\le\ln(a)+\frac {\frac za-1}{\sqrt{1+\frac za-1}}$$
or for $x\ge a$ and $a>0$ :
$$\ln(x)\le\ln(a)+\frac{x-a}{\sqrt{ax}}$$
(as obtained by coffeemath...)
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