How find this minimun of this $\frac{xy}{x^5+xy+y^5}+\frac{yz}{y^5+yz+z^5}+\frac{xz}{x^5+xz+z^5}$ let $x,y,z>0$ and such
$$x+y+z=1$$
Find minimum of the 
$$\dfrac{xy}{x^5+xy+y^5}+\dfrac{yz}{y^5+yz+z^5}+\dfrac{xz}{x^5+xz+z^5}$$
I have find this maximum
note
$$\dfrac{xy}{x^5+xy+y^5}\le -243\dfrac{x+y}{841}+\dfrac{23031}{1682}$$
see
http://www.wolframalpha.com/input/?i=xy%2F%28x%5E5%2Bxy%2By%5E5%29%2B243*%28x%2By%29%2F841-23031%2F1682
But for minimum value,I can't,(maybe use AM-GM) Thank you,

The max is $\dfrac{81}{29}$ and min is $\dfrac{.25}{2*(0.5)^5+.25}$, a simple method is let two varies equal$(x=y)$ and have a function of $x$, see graphic below:
 
edit :to prove max, consider $xy \le \dfrac{x^2+y^2}{2}, x^5+y^5 \ge 2(\dfrac{x^2+y^2}{2})^{\frac{5}{2}},\dfrac{xy}{x^5+xy+y^5}\le \dfrac{x^2+y^2}{x^2+y^2+4(\dfrac{x^2+y^2}{2})^{\frac{5}{2}}}=\dfrac{p}{p+4(\dfrac{p}{2})^{\frac{5}{2}}}=f(p)$
$f(p)$ is concave and mono decreasing  function and $x^2+y^2+z^2 \ge \dfrac{1}{3}$ ,so we can quickly know the max will be got when $x=y=z$
