Upper Bound of Logarithm

$\ln(x)=-\ln\left(1-(1-\frac{1}{x})\right)$

Since $x\geq 2$ then $\left|1-\frac{1}{x}\right|<1$

Using the Taylor series of the logarithm: $$\ln(1-t)=-\sum\limits_{k=1}^{\infty}\frac{t^n}{n}$$ with $t=1-\frac{1}{x}$ in this case, we obtain: $$\ln x\leq 1-\frac{1}{x}$$ $$\ln x\leq \left(1-\frac{1}{x}\right)+\frac{\left(1-\frac{1}{x}\right)^2}{2}+\frac{\left(1-\frac{1}{x}\right)^3}{3}$$ You can stop at any add degree. In general: $$\ln x\leq\sum\limits_{k=1}^{2n+1}\frac{\left(1-\frac{1}{x}\right)^k}{k}$$

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.

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...)