Can anyone prove this function is concave? Can anyone help me prove the following function $$f(x) = \frac{ax}{1-x^a}-\frac{x}{1-x}$$ is concave for any parameter $a \geq 1$ when $0<x<1$ ?
I have tried to prove its second derivative is negative, but it becomes another difficult problem.
 A: AmerYR calculated $$f''(x)=\frac{a^2x^{a-1}\left(1+a+x^a(a-1)\right)}{(1-x^a)^3} - \frac{2}{(1-x)^3} \, .$$
As you noticed $f''(x)=0$ at $a=1$. So it suffices to show that $\partial_a f''(x) \leq 0$. In the comments I gave the derivative with respect to $a$ by $${\frac { \left(  \left(  \left( {a}^{2}-a \right) \ln  \left( x  \right) -3\,a+2 \right) {x}^{3\,a-1}+ \left( 4\,{a}^{2}\ln  \left( x  \right) -4 \right) {x}^{2\,a-1}+{x}^{a-1} \left(  \left( {a}^{2}+a  \right) \ln  \left( x \right) +3\,a+2 \right)  \right) a}{ \left( 1- {x}^{a} \right) ^{4}}} \, .$$ Since $\frac{(1-x^a)^4}{a \, x^{a-1}}>0$ we can multiply by this and obtain the objective $$\left(  a\left( a-1 \right) {x}^{2\,a}+4\,{a}^{2}{x}^{a}+{a}^{2}+a \right) \ln  \left( x \right) + \left( -3\,a+2 \right) {x}^{2\,a}-4 \,{x}^{a}+3\,a+2 \stackrel{!}{\leq} 0 \, .\tag{0}$$
We now need to get rid of the logarithm in order to get a manifest signature. I leave it to you to check that this expression goes to $-\infty$ as $x\rightarrow 0$, while it vanishes at $x=1$. Taking the derivative with respect to $x$ and dividing the resulting expression by $a \, x^{a-1}>0$ yields $$2\,a \left(  \left( a-1 \right) {x}^{a}+2\,a \right) \ln  \left( x \right) + \left( a+1 \right) {x}^{-a}+ \left( -5\,a+3 \right) {x}^{a} +4\,a-4 \stackrel{!}{\geq}0 \tag{1} \, .$$
Now rinse and repeat. Check the limiting cases (they are $+\infty$ and $0$), derive by $x$ and divide by $a\,x^{a-1}$ to obtain $$2a\left( a-1 \right) \ln  \left( x \right) - \left( a+1 \right) {x}^{-2\,a}+4\,{x}^{-a}a-3\,a+1 \stackrel{!}{\leq} 0 \, .\tag{2}$$ Almost there; The limiting cases are manifestly $-\infty$ and $0$ respectively and so after deriving with respect to $x$ and multiplying by $\frac{x^{2a+1}}{2a}>0$ this becomes $$\left( a-1 \right) {x}^{2\,a}-2\,a\,{x}^{a}+a+1 \stackrel{!}{\geq} 0 \, .\tag{3}$$ One more time: At $x=0$ this is $a+1>0$ and at $x=1$ it vanishes again. Deriving with respect to $x$ and dividing by $2a\, x^{a-1}>0$ we have $$(a-1)x^a - a \stackrel{!}{\leq} 0 \, .\tag{4}$$
A: $f(x) = \frac{ax}{1-x^a} - \frac{x}{1-x} $ , $a\geq 1 , 0<x<1$ 
$$f'(x) =\frac{a}{1-x^a} + \frac{ax(ax^{a-1})}{(1-x^a)^2} - \frac{1}{(1-x)^2} $$
$$f'(x) = \frac{a}{1-x^a} + \frac{a^2 x^a}{(1-x^a)^2} - \frac{1}{(1-x)^2}$$
$$f''(x) = \frac{a^2x^{a-1}}{(1-x^a)^2} + \frac{a^3x^{a-1}}{(1-x^a)^2} + \frac{2a^3x^{2a-1}}{(1-x^a)^3}- \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)}{(1-x^a)^2}+  \frac{2a^3x^{2a-1}}{(1-x^a)^3}- \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)-a^2x^{2a-1}(1+a)+2a^3x^{2a-1}}{(1-x^a)^3} - \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)-a^2x^{2a-1}+a^3x^{2a-1}}{(1-x^a)^3} - \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a-x^a+ax^a)}{(1-x^a)^3} -\frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a+x^a(a-1))}{(1-x^a)^3} -\frac{2}{(1-x)^3}$$
 we need to show it is negative for the given conditions. For a fixed $x$ we can see that the first fraction goes to zero as $a$ increases because $0<x<1$. Can you continue? 
