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.
 A: Something different than brute force would be the usage of Schur's inequality and your transformation of $x^4+y^4=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$
$$\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$$
$$(x+y+z)\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
$$\sum_{cyc}(x^4+y^4+\frac{(x^4+y^4)z}{x+y})\le 3(x^4+y^4+z^4)$$
$$\sum_{cyc}((x^3+y^3)z-xyz(x+y)+\frac{2x^2y^2z}{x+y})\le x^4+y^4+z^4$$
$$\sum_{cyc}(x^3z+y^3z)+2xyz\sum_{cyc}\frac{xy}{x+y}\le x^4+y^4+z^4+2xyz(x+y+z)$$
Bcs of the AM-HM inequality we have $x+y+z\ge \frac{2xy}{x+y}+\frac{2yz}{y+z}+\frac{2zx}{z+x}$ and the other part we obtain from Shurs inequality for $t=2$:
$x^2(x-y)(x-z)+y^2(y-z)(y-x)+z^2(z-x)(z-y)\ge0$. Add those two together and we get our desired inequality.
A: 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
A: Let $x\geq y\geq z$. Hence,
$$y^2\left(\frac{3(x^4+y^4+z^4)}{x+y+z}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=y^2\sum\limits_{cyc}\left(\frac{3(x^4+y^4)}{2(x+y+z)}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=$$
$$=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(x+y-2z)}{2(x+y+z)(x+y)}=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(y-z-(z-x))}{2(x+y+z)(x+y)}=$$
$$=\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}(x-y)\left(\frac{x^4+z^4}{x+z}-\frac{y^4+z^4}{y+z}\right)=$$
$$=\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}\frac{(x-y)^2(xy(x^2+xy+y^2)+z(x+y)(x^2+y^2)-z^4)}{(x+z)(y+z)}\geq$$
$$\geq\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}\frac{(x-y)^2(z(x^3+y^3)-z^4)}{(x+z)(y+z)}=$$
$$=\frac{y^2}{2(x+y+z)\prod\limits_{cyc}(x+y)}\sum\limits_{cyc}(x-y)^2z(x^3+y^3-z^3)(x+y)\geq$$
$$\geq\frac{y^2\left((x-z)^2y(x^3-y^3)(x+z)+(y-z)^2x(y^3-x^3)(y+z)\right)}{2(x+y+z)\prod\limits_{cyc}(x+y)}\geq$$
$$\geq\frac{x^2(y-z)^2y(x^3-y^3)(x+z)+y^2(y-z)^2x(y^3-x^3)(y+z)}{2(x+y+z)\prod\limits_{cyc}(x+y)}=$$
$$=\frac{xy(y-z)^2(x^3-y^3)(x(x+z)-y(y+z))}{2(x+y+z)\prod\limits_{cyc}(x+y)}=$$
$$=\frac{xy(y-z)^2(x-y)^2(x^2+xy+y^2)}{2\prod\limits_{cyc}(x+y)}\geq0.$$
Done!
