Find $k$ such that $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}$ for $xyz=1$. 
Find all $k\in\mathbb{R}^+$ such that for all $xyz=1$, $x,y,z\in\mathbb{R}^+$, we
  have
  $$\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}.$$

If we set $$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a},$$ where $a,b,c\in\mathbb{R}^+$, then the inequality is equivalent to $$\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\geq\frac{3}{2^k}.$$ For $k=1$ this is obviously false, as we take $b\to0$ and $c\to\infty$. In an olympiad test I saw that the case $k=\sqrt3$ is true (though I do not know how to prove that). I then thought that maybe this is true for all $k>1$. Am I correct and how do I prove it?

I think the answer is $k_{min}=\log_23$.
Let $x=e^a$, $y=e^b$, $z=e^c$ and $f(x)=\frac{1}{(1+e^x)^k}$.
Hence, $a+b+c=0$ and we need to prove that 
$$\sum_{cyc}f(a)\geq3f\left(\frac{a+b+c}{3}\right)$$
But $f''(x)=\frac{2ke^x(ke^x-1)}{(1+e^x)^{k+2}}\geq0$ for all $k\geq1$ and $x\geq0$.
Thus, by Vasc's RCF Theorem for all $k\geq1$ it's enough to prove our inequality
for $y=x$ and $z=\frac{1}{x^2}$, which gives $\frac{2}{(1+x)^k}+\frac{1}{\left(1+\frac{1}{x^2}\right)^k}\geq\frac{3}{2^k}$
and for $x\rightarrow+\infty$ we get $k\geq\log_23$.
