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.

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.
