Determine for which $x$ and $\alpha$ holds $\frac{x^\alpha}{(x^\alpha+(1-x)^\alpha)^\frac{1}{\alpha}}>A$ Recently, I posted a question (see here), but didn't recieve a complete answer. Therefore, I would like to simplify it, and rewrite the problem as the following: for which values of $x,\alpha,\beta$ holds the inequality 
$$\frac{x^\alpha}{(x^\alpha+(1-x)^\alpha)^\frac{1}{\alpha}}>A,$$
where $x,\alpha\in[0,1]$ and $A$ is some constant.
 A: Let's assume throughout that $A <1$, and let's call your function
$$
f(x,\alpha) = \frac{x^\alpha}{(x^\alpha+(1-x)^\alpha)^\frac{1}{\alpha}}
$$
Observe that for all $\alpha >0$, $0 \le f(x,\alpha) \le 1$ where
$$
\lim_{x \to 0} f(x,\alpha) = 0 
$$
and
$$
\lim_{x \to 1} f(x,\alpha) = 1 
$$
, and nowhere else is  $f(x,\alpha) =0$ or $f(x,\alpha) =1$. This  means that for $f(x,\alpha) \ge A$ to hold, you will always have to require some $x(A, \alpha) > 0$. Some numerics show that $f(x,\alpha)$ is rising monotonously with $x$ for all $\alpha \sim > 0.2792$. So in these cases, the inequality holds for all $x \ge x_0$ where $x_0$ is the only solution to $f(x_0,\alpha) = A$ - you get that solution only numerically.
For all  $\alpha < \sim 0.2792$, we have that  $f(x,\alpha)$ has a relative maximum $f_+$ at some value $x_+$ and a  relative minimum $f_- < f_+$ at some value $x_- > x_+$.  
Then, if $A < f_-$, you have that the inequality holds for all $x \ge x_0$ where $x_0$ is the smallest solution to $f(x_0,\alpha) = A$.
If $f_- < A < f_+$, you have that the inequality holds in two intervals of $x$: $[x_0,  x_1]$ and $[x_2,  1]$ where $x_0 < x_+ < x_1 < x_- < x_2$.
If $f_+< A$, you have that the inequality holds for all $x \ge x_0$ where $x_0$ is the only solution to $f(x_0,\alpha) = A$.
Since you get all of these values only numerically, the best you can obtain is the general behaviour of the function $f(x,\alpha)$ which I described.
