Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$ 
Let $x,y,z\ge 0$. Prove or disprove
  $$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$

This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \sum_{cyc}\dfrac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
and
$$(x^4+y^4)=(x+y)(x^3+y^3)-xy(x^2+y^2)=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$$
which is equivalent to 
$$\sum_{cyc}\left((x^3+y^3)-xy(x+y)+\dfrac{2x^2y^2}{x+y}\right)\le 3(x^4+y^4+z^4)$$
or
$$2\sum_{cyc}x^3+2\sum_{cyc}\dfrac{x^2y^2}{x+y}\le 3\sum_{cyc}(x^4+xy(x+y))$$
and now I'm stuck.

we have to show that $$\frac{3(x^4+y^4+z^4)}{x+y+z}-\frac{x^4+y^4}{x+y}-\frac{z^4+y^4}{z+y}-\frac{z^4+x^4}{x+z}\geq 0$$
assuming we have $$x=\min(x,y,z)$$ then we set
$$y=x+u,z=x+u+v$$ and we have after some algebra
$$\left( 12\,{u}^{2}+12\,uv+12\,{v}^{2} \right) {x}^{5}+ \left( 30\,{u}
^{3}+45\,{u}^{2}v+75\,u{v}^{2}+30\,{v}^{3} \right) {x}^{4}+ \left( 28
\,{u}^{4}+56\,{u}^{3}v+144\,{u}^{2}{v}^{2}+116\,u{v}^{3}+28\,{v}^{4}
 \right) {x}^{3}+ \left( 12\,{u}^{5}+30\,{u}^{4}v+120\,{u}^{3}{v}^{2}+
150\,{u}^{2}{v}^{3}+72\,u{v}^{4}+12\,{v}^{5} \right) {x}^{2}+ \left( 2
\,{u}^{6}+6\,{u}^{5}v+45\,{u}^{4}{v}^{2}+80\,{u}^{3}{v}^{3}+57\,{u}^{2
}{v}^{4}+18\,u{v}^{5}+2\,{v}^{6} \right) x+6\,{u}^{5}{v}^{2}+15\,{u}^{
4}{v}^{3}+14\,{u}^{3}{v}^{4}+6\,{u}^{2}{v}^{5}+u{v}^{6}
\geq 0$$ which is true.
the other cases are analogously
