I am studying the following function, defined in the support $[0,a]$:
$f(x)=(1+a)^{\frac{1+a}{a+x}}(1+x)^{\frac{1+x}{a+x}}$
In particular I would like to prove that the function $f(x)$ is concave in $x$, so I just need to prove that the function has only a critical point, but I cannot find it (or them) in any analytical way.
Write $$\ln f(x) = \frac{h(x)+h(a)}{x+a},$$ where $h(x) = (1+x)\ln(1+x),\ x\ge 0$. Then, $$\frac{f'(x)}{f(x)}=\frac{h'(x)(x+a)-h(x)-h(a)}{(x+a)^2}=\frac{g(x)}{(x+a)^2},$$ where $g(x)=h'(x)(x+a)-h(x)-h(a)=x+a+(a-1)\ln(1+x)-(a+1)\ln(1+a)$. Note that $g(0)=a-(a+1)\ln(1+a)\le 0$, since the function $x-(1+x)\ln(1+x)$ is decreasing for $x>0$ (check!). Also, $g(a)=2(a-\ln(1+a))>0$, since $x+1<e^x,\forall x>0 $. Hence, $f'(0)\le 0$, and $f'(a)>0$, since $f(x)>0,\ \forall x\ge 0$. Hence, one cannot have $f''(x)\le 0,\ \forall x\in [0,a]$, meaning that the function is not concave in $[0,a]$.
