How can we find minimum of $f(x,y,z)?$ Let $k\in\mathbb{N}$ and $x,y$ and $z$ are positive real number such that $x+y+z=1$. How can we find minimum of $f(x,y,z)$ where
$$f(x,y,z)=\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}}$$
I think (if I am right) that we can use chebyshev inequality but I am not sure.

Chebyshev Inequality 
If $\{a_i\}_{i=1}^{n}$ and $\{b_i\}_{i=1}^{n}$ be two
  non-increasing sequences then
  $$n(a_1b_1+a_2b_2+\cdots+a_nb_n)\ge(a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots+b_n)$$
   Or
  if $\{a_i\}_{i=1}^{n}$ be non-increasing sequence and  $\{b_i\}_{i=1}^{n}$ be a 
  non-decreasing sequence then
  $$n(a_1b_1+a_2b_2+\cdots+a_nb_n)\le(a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots+b_n)$$

Thanks for your hints, references or answers.

This is not the complete solution. In here, I give the range of values in which $f(x,y,z)$ lies. First, note that since $x\geq 0,~y\geq 0,~z\geq 0$ and $x+y+z=1$, we must have $0\leq x,y,z\leq 1$. Thus, for any $k\in \mathbb{N}$, we have
\begin{eqnarray}
x^{k+1}< x^{k},~y^{k+1}< y^{k},~\text{and}~z^{k+1}< z^{k}.
\end{eqnarray}
Thus, we have
\begin{align}
f(x,y,z)&> \frac{x^{k+2}+y^{k+2}+z^{k+2}}{x^{k}+y^{k}+z^{k}}.
\end{align}
Without loss of generality, assume $x\geq y\geq z$. Then, applying the first Chebyshev's inequality with $a_{1}=x^{2},~a_{2}=y^{2},~a_{3}=z^{2},~b_{1}=x^{k},~b_{2}=y^{k},~b_{3}=z^{k}$, we get
\begin{align}
3(x^{k+2}+y^{k+2}+z^{k+2})&\geq ({x^{2}+y^{2}+z^{2}})({x^{k}+y^{k}+z^{k}}).
\end{align}
We thus have
\begin{align}
f(x,y,z)&> \frac{1}{3}(x^{2}+y^{2}+z^{2})\\
        &\stackrel{(a)}{\geq}\frac{1}{9},
\end{align}
where, in writing (a) above, I have used the solution $(x^{*},y^{*},z^{*})=(1/3,1/3,1/3)$ to the following optimization problem:
\begin{eqnarray}
(x^{*},y^{*},z^{*})=\arg\min ~~(x^{2}+y^{2}+z^{2})\\
\text{subject to}~~~~~x+y+z=1,~x\geq 0,~y\geq 0,~z\geq 0.
\end{eqnarray}
Thus, $f(x,y,z)>\frac{1}{9}$, and by inspection, we find $f(x,y,z)\leq 1$. Thus, $f(x,y,z)\in (\frac{1}{9},1]$.
