Prove that $\sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$.

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

Here's what I did.

$$\large \sum_{cyc}\dfrac{a}{a + b^4 + c^4}$$

$$\le \sum_{cyc}\dfrac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \dfrac{1}{(a + b + c)^2}\sum_{cyc}\dfrac{ab^2 + ac^2 + 1}{(bc)^2} \le \dfrac{1}{9}\sum_{cyc}\left(\dfrac{a}{b^2} + \dfrac{a}{c^2} + a^2\right)$$

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.)

$$\large \sum_{cyc}\dfrac{a}{a + b^4 + c^4}$$

$$= 3 - \sum_{cyc}\dfrac{b^4 + c^4}{a + b^4 + c^4} \le 3 - 4\sum_{cyc}\dfrac{a^4}{b + c + c^4 + 2a^4 + b^4} = 2\sum_{cyc}\dfrac{b + c + b^4 + c^4}{b + c + c^4 + 2a^4 + b^4} - 3$$

...and again...

$$\large \sum_{cyc}\dfrac{a}{a + b^4 + c^4}$$

$$= 3 - \sum_{cyc}\dfrac{b^4 + c^4}{a + b^4 + c^4} \le 3 - \dfrac{1}{2}\sum_{cyc}\dfrac{(b^2 + c^2)^2}{a + b^4 + c^4} = \sum_{cyc}\dfrac{2(a - bc) + (b^4 + c^4)}{a + b^4 + c^4}$$

$$\le \sum_{cyc}\dfrac{1}{2(a - bc) + (b^4 + c^4)}\dfrac{[2(a - bc) + (b^4 + c^4)]^2}{a + 2b^2c^2}$$

$$= \dfrac{1}{2}\sum_{cyc}\dfrac{(a - bc + b^4)^2 + (a - bc + c^4)^2}{[2(a - bc) + (b^4 + c^4)](a + 2b^2c^2)}$$

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.
By C-S $$\sum_{cyc}\frac{a}{a+b^4+c^4}-1=\sum_{cyc}\frac{a(a^3+2)}{(a+b^4+c^4)(a^3+1+1)}-1\leq$$ $$\leq\sum_{cyc}\frac{a^4+2a}{(a^2+b^2+c^2)^2}-1 =-\frac{2\sum\limits_{cyc}(a^2b^2-a^2bc)}{(a^2+b^2+c^2)^2}=-\frac{\sum\limits_{cyc}c^2(a-b)^2}{(a^2+b^2+c^2)^2}\leq0.$$