$\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 $\alpha > 0$?
 A: Define
$$f(\alpha) = \alpha (1+\alpha/2)^{-1} - \log(1+\alpha)$$
$$\implies f'(\alpha) = \frac{-\alpha^2}{(\alpha+1)(\alpha+2)^2}$$
This is strictly negative for $\alpha>0$, so it is decreasing. Since
$$f(0) = 0(1+0/2)^{-1} - \log(1+0) = 0$$
$$\implies f(\alpha) = \alpha (1+\alpha/2)^{-1} - \log(1+\alpha) < 0 \quad \text{for} \quad \alpha>0$$
This proves the result. $\blacksquare$
A: This is the same as
$a  < (1+a/2)\log(1+a)
$.
Since
$\log(1+a) < a-a^2/2
$,
$(1+a/2)\log(1+a)
<(1+a/2)(a-a^2/2)
=a(1+a/2)(1-a/2)
=a(1-a^2/4)
< a
$.
A: 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
$$
I > \int_0^\alpha \frac{dx}{1+\alpha/2} = \frac{\alpha}{1+\alpha/2}.
$$
