Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$ 
If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$

Here's what I did.
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \frac{1}{(a + b + c)^2}\sum_{cyc}\frac{ab^2 + ac^2 + 1}{(bc)^2} \le \frac{1}{9}\sum_{cyc}\left(\frac{a}{b^2} + \frac{a}{c^2} + a^2\right)$$
And this is where I gave up. 
Thanks for reading this. Even more thanks if you can solve the problem.
(Oh wait, I did try again.)
$$\sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - 4\sum_{cyc}\frac{a^4}{b + c + c^4 + 2a^4 + b^4} = 2\sum_{cyc}\frac{b + c + b^4 + c^4}{b + c + c^4 + 2a^4 + b^4} - 3$$
...and again...
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - \frac{1}{2}\sum_{cyc}\frac{(b^2 + c^2)^2}{a + b^4 + c^4} = \sum_{cyc}\frac{2(a - bc) + (b^4 + c^4)}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{1}{2(a - bc) + (b^4 + c^4)}\frac{[2(a - bc) + (b^4 + c^4)]^2}{a + 2b^2c^2}$$
$$= \frac{1}{2}\sum_{cyc}\dfrac{(a - bc + b^4)^2 + (a - bc + c^4)^2}{[2(a - bc) + (b^4 + c^4)](a + 2b^2c^2)}$$
And I need help. I would be grateful if you could help with this problem.

We can use rearrangement inequality to obtain$$
b^4+c^4\ge b^3c+bc^3=bc(b^2+c^2)=\frac{b^2+c^2}a.
$$ This implies
$$\begin{align*}
\sum_{cyc}\frac{a}{a+b^4+c^4}&\le\sum_{cyc}\frac{a}{a+\frac{b^2+c^2}a}\\&=\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}\\&=\frac{\sum_{cyc}a^2}{a^2+b^2+c^2}\\&=1,
\end{align*}$$ as desired.
