Intuition for an upper bound on $\ln(x)$ for $x \geq 1$ Today I came across the following bound. For $x \geq 1$,
$$\ln(x) \leq \frac{x^2 - 1}{2x}.$$
(it provides a nice way of showing that $\ln(2) \leq 3/4$ when you don't have a calculator and you don't want to check the first 8 or so terms of the Taylor expansion for $\ln(x + 1)$)
It is fairly easy to prove using some pre-calculus tools, but can anyone give an idea of how someone might come up with a bound of this form?
 A: Consider the function $$f(x)=\ln(x)-\dfrac{x^2 - 1}{2x}$$ on $[1,\infty).$ This function is smooth and $\dfrac{df}{dx}=-\dfrac{(x-1)^2}{2x^2}$ and $\dfrac{d^2f}{dx^2}=\dfrac{1-x}{x^3}.$ Thus this function is strictly decreasing and concave down. As $f(1)=0,$ function is negative for all $x\gt 1.$
EDIT:
Best way to come up with these type of inequalities is mean value theorems. Also $\ln x$ has a very slow growth rate (look at the graph), there for any fast growing function would exceed $\ln x$ at some point. In your case that function is $\dfrac12\left(x-\dfrac1x\right)$ and the crossing point is $x=1.$
A: By the Cauchy-Schwarz inequality, for any $x\geq 1$ we have
$$ \log(x)=\int_{1}^{x}\frac{du}{u}\leq\sqrt{\int_{1}^{x}du\int_{1}^{x}\frac{du}{u^2}} = \sqrt{x}-\frac{1}{\sqrt{x}} $$
which is sharper than the wanted inequality.

Newton's method then allows to state
$$ \log(x)\leq \sqrt{x}-\frac{1}{\sqrt{x}}-1+\frac{x}{\exp\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)} $$
and so on. Other accurate bounds can be derived from the (generalized) Shafer-Fink inequality, for instance.

Anyway, one of the most efficient ways for approximating $\log 2$ is to exploit Beuker-like integrals:
$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x}\,dx = -\frac{621}{56}+16\log 2$$
and since the LHS is bounded between $0$ and $\frac{1}{4^4}$, we have $\log(2)\leq\frac{39}{56}$ which is both simple and pretty accurate.
