<s>[INST] 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.
 [/INST] $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? 
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