<s>[INST] Proof regarding the minimum of a function I want to show that the function 
$$f(x)= \frac{2x}{1+(\frac{1}{k-x})^a} +  \frac{k-2x}{1+2\cdot(\frac{1}{k-x})^a+ (\frac{1}{k-2x})^a}$$
with 
$$ x \in [0,\frac{k}{2}], a \in [0,\infty], k \in [0,1] $$
has the minimum 
$$min(f(x)) = \begin{cases}
\frac{k}{1+3\cdot(\frac{1}{k})^a},\, a\leq \frac{log(3)}{log(2)}
\\ 
\frac{k}{1+(\frac{1}{k/2})^a} ,\, else
\end{cases}$$
for all permited values of the parameters $a,k$. The minima can be obtained by setting either $x=0$ or $x=\frac{k}{2}$. I checked that this is in fact the minimum by simulating the function for all permitted parameter values.
My initial idea was to use the generalized version of the Binomial theorem 
on the $1/(k-x)^a$ terms, but this only worked for $a\leq2$ because the elements of the series obtained from the theorem are not shrinking in absolute value. 
Another idea was to show that every critical point of the function is never a global minimum, but I cannot find a solution for $f'(x)=0$. 
Any ideas would be greatly appreciated. Thanks in advance.
 [/INST] Update
We can also prove the case when $a > \frac{11}{4}$. 
We will use Fact 1 and 2 whose proof is simple and thus omitted.
Fact 1: If $a > \frac{11}{4}$, we have
$$\frac{2x}{1+(\frac{1}{k-x})^a} \ge \frac{k}{1 + (\frac{2}{k})^a}$$
for $\frac{k}{4} \le x \le \frac{k}{2}$. 
Fact 2: If $a > \frac{11}{4}$, we have
$$\frac{k^2}{k + 2(k-x)^{1-a} + (k-2x)^{1-a}} \ge \frac{k}{1+(\frac{2}{k})^a}$$
for $0\le x \le \frac{k}{4}$.
Now let us proceed.
If $0\le x\le \frac{k}{4}$, using the Cauchy-Bunyakovsky-Schwarz inequality and Fact 2, we have
$$f(x) \ge \frac{k^2}{k + 2(k-x)^{1-a} + (k-2x)^{1-a}} \ge  \frac{k}{1+(\frac{2}{k})^a}.$$
If $\frac{k}{4} \le x \le \frac{k}{2}$, using Fact 1, we have
$$f(x) \ge \frac{2x}{1+(\frac{1}{k-x})^a} \ge \frac{k}{1 + (\frac{2}{k})^a}.$$
Previously written
Partial answer
I prove the case when $0 < a < 1$. (The case $a=1$ is easy.)
Let
\begin{align}
g(x) &= 2x \Big(1 + \Big(\frac{1}{k-x}\Big)^a\Big)
+ (k-2x)\Big(1 + 2\Big(\frac{1}{k-x}\Big)^a + \Big(\frac{1}{k-2x}\Big)^a\Big)\\
&= k + 2(k-x)^{1-a} + (k-2x)^{1-a}.
\end{align}
Using the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$f(x) \ge \frac{(2x + k-2x)^2}{g(x)} = \frac{k^2}{g(x)}.$$
If $0 < a < 1$, then $g(x)$ is strictly decreasing for $x\in [0, k/2]$.
We have $f(x) \ge \frac{k^2}{g(0)} = \frac{k^2}{k+ 3k^{1-a}} = f(0)$.We are done.
</s>