$\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$ How does one prove that $\alpha (1+\alpha/2)^{-1} < \log(1+\alpha) $ for $\alpha > 0$?

By changing variable $x \rightarrow a-x$, we have
$$
I = \int_0^\alpha \frac{dx}{1+x} = \int_0^\alpha \frac{dx}{1+\alpha-x}.
$$
So, by averaging the two expressions, we get
$$
I = \int_0^\alpha \frac{1}{2} \left( \frac{1}{1+x} + \frac{1}{1+\alpha-x} \right) \, dx
$$
But by the convexity of $1/(1+x)$, we have
$$
\frac{1}{2} \left( \frac{1}{1+x} + \frac{1}{1+\alpha-x} \right)
\ge \frac{1}{1+(x + \alpha-x)/2} = \frac{1}{1+\alpha/2}.
$$
The equality is impossible when $x\ne\alpha/2$.  So
$$
I > \int_0^\alpha \frac{dx}{1+\alpha/2} = \frac{\alpha}{1+\alpha/2}.
$$
