Who can tell me how to prove that $f(x) = \frac{x}{\log_2(1+x)}$ is concave? Thanks a lot in advance.
Take the second derivative: $f^{''}(x) = \frac{-\ln2*(x+2)*\big(\log_2(1+x)\big)^2+2x\log_2(1+x)}{\big(\ln2*(1+x)\log_2(1+x))^2\big)^2}$.
How to decide the second derivative is a negative value?
The second derivative, ignoring the $_2$ which doesn't matter, is $\dfrac{2 x - (x + 2) \ln(1 + x)}{(1 + x)^2 \ln^3(1 + x)} $ so we want to show that $g(x) =2 x - (x + 2) \ln(1 + x) \lt 0$.
$g(0) = 0$. $g(1) =2-3\ln(2) \approx -.08 $, $g'(x) =2-\ln(1+x)-\dfrac{x+2}{1+x} =-\ln(1+x)+\dfrac{2+2x-(x+2)}{1+x} =-\ln(1+x)+\dfrac{x}{1+x} $ but $\ln(1+x) =\int_0^x \dfrac{dt}{1+t} \ge\int_0^x \dfrac{dt}{1+x} =\dfrac{x}{1+x} $ so $g'(x) \le 0$ and, for large $x$, $\ln(1+x) \gt 2 $ for $x > e^2-1 $ so $g(x) \lt 2x-2(x+2) =-4 $ for $x > e^2-1$.
Therefore $g(x) < 0$ for $x > 0$ and the function is concave.