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?

As shown below, an easy partial result is that $k_{\text{min}} \ge 1$.

Let $k$ be fixed with $0 <  k < 1$. 

WIth the substitutions
$$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a}$$
let $a = b = 1$, and let $c \to 0^{+}$. Then
\begin{align*}
\;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt]
&\to \;\frac{1}{2^k} + 1 + 0\\[6pt]
&=\;\frac{1 + 2^k}{2^k}\\[6pt]
&<\;\frac{1 + 2}{2^k}\\[6pt]
&=\;\frac{3}{2^k}\\[6pt]
\end{align*}
It follows that $k_{\text{min}} \ge 1$, as claimed.

However, as shown below, the reasoning by which the OP (Yuxiao Xie) excluded the value $k=1$ is not correct.

Suppose $k = 1$. Then as$\;\,b \to 0^{+}$ and $\,c \to \infty$,
\begin{align*}
\;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt]
&\to\;1+ 0 + 1\\[6pt]
&=\;2\\[6pt]
&>\;\frac{3}{2}\\[6pt]
\end{align*}
which doesn't break the required inequality, hence the value $k=1$ is not automatically excluded.

In fact, I suspect that $k_{\text{min}} = 1$.

Update: It's not true, as I had previously thought, that $k_{\text{min}} = 1$.

As a simple counterexample, let $(x,y,z) = \left(\frac{1}{8},2,4\right)$. Then 
$$\frac{1}{(1+x)}+\frac{1}{(1+y)}+\frac{1}{(1+z)} = \frac{64}{45} < \frac{3}{2}$$
