If $x,y,z>0$, then $\frac{x}{4x+4y+z} + \frac{y}{4y+4z+x}+\frac{z}{4z+4x+y}\le \frac13$ I have a prove of the following inequality that depends upon somewhat messy algebra. 
I would like to learn how to prove it in a more elegant way. 

For positive numbers: $$\frac{x}{4x+4y+z} + \frac{y}{4y+4z+x}+\frac{z}{4z+4x+y}\le \frac13$$

 A: Wlog, we may assume $x+y+z = 3$. So, the inequality, $$\frac{x}{4x+4y+z} + \frac{y}{4y+4z+x}+\frac{z}{4z+4x+y}\le \frac13$$
$$\iff \sum\limits_{cyc} \frac{x}{4-z} \le 1 \iff \sum\limits_{cyc} x(4-x)(4-y) \le \prod\limits_{cyc} (4-x)$$
$$\iff x^2y+y^2z+z^2x +xyz \le 4$$
Assume, now wlog that, $max\{x,z\} \ge y \ge \min\{x,z\}$.
Then, $$z(x-y)(z-y) \le 0 \iff y^2z+xz^2 \le xyz + yz^2$$
So, it remains to prove that, $$x^2y + yz^2 +2xyz \le 4 \iff y(x+z)^2 \le 4 \iff y(3-y)^2 \le \frac12\left(\frac{6-2y+2y}{3}\right)^3$$
by AM-GM inequality.
Edit: Well there is another cute little observation about the way OP and David H (in the comment), writes the inequality, that might reduce the calculation effort as: $$0 \le 4(x^3+y^3+z^3) + 12(x^2z+xy^2+yz^2) - 15(x^2y+xz^2+y^2z) - 3xyz$$
$$\iff 12\left(\sum\limits_{cyc} x^2y - \sum\limits_{cyc} xy^2\right) \le 3\left(\sum\limits_{cyc} x^3 - \sum\limits_{cyc}x^2y\right) + \left(\sum\limits_{cyc} x^3 - 3xyz\right)$$
$$\iff 12(x-y)(x-z)(y-z) \le \sum\limits_{cyc} (2z+z)(y-z)^2+\frac12(x+y+z)\sum\limits_{cyc} (y-z)^2$$
$$\iff 12(x-y)(x-z)(y-z) \le \sum\limits_{cyc} \left(2y+z+\frac12(x+y+z)\right)(y-z)^2$$
Observing that $(x-y),(y-z)$ and $(z-x)$ remain invariant under the transformations: 
$x \mapsto x-t$ , $y \mapsto y-t$ and $z \mapsto z-t$
It suffices to prove the inequality for the case where terms in RHS is minimized, i.e., when we let $t = \min\{x,y,z\} = 0$.
Wlog, if $z=\min\{x,y,z\}$, then it reduces $x^2y+y^2z+z^2x +xyz \le 4$ to proving $x^2y \le 4$ under the condition $x+y=3$, which follows from Am-Gm as usual.
A: We need to prove that
$$\sum_{cyc}\frac{x^2+xy+xz}{4x+4y+z}\leq\frac{x+y+z}{3}$$ or
$$\sum_{cyc}\left(\frac{x^2+xy+xz}{4x+4y+z}-\frac{x}{4}\right)\leq\frac{x+y+z}{3}-\frac{x+y+z}{4}$$ or
$$\sum_{cyc}\frac{xz}{4x+4y+z}\leq\frac{x+y+z}{9},$$
which is true by C-S:
$$\sum_{cyc}\frac{xz}{4x+4y+z}=\sum_{cyc}\frac{xz}{2(2x+y)+2y+z}\leq$$
$$\leq\sum_{cyc}\frac{xz}{9}\left(\frac{2^2}{2(2x+y)}+\frac{1^2}{2y+z}\right)=
\frac{1}{9}\sum_{cyc}\left(\frac{2xz}{2x+y}+\frac{xz}{2y+z}\right)=$$
$$=\frac{1}{9}\sum_{cyc}\left(\frac{2xy}{2y+z}+\frac{xz}{2y+z}\right)=\frac{x+y+z}{9}.$$
Done!
