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If $a,b,c \in \mathbb{R+}$ and $abc=1$ Prove That

$$S=\sum \frac{a}{a+b^4+c^4} \le 1$$

My approach:

we have $$S=\sum \frac{\frac{1}{bc}}{\frac{1}{bc}+b^4+c^4}=\sum \frac{1}{1+b^5c+bc^5}$$

Now by $AM \ge GM$ we have

$$\frac{1+b^5c+bc^5}{3} \ge b^2c^2$$ $\implies$

$$\frac{1}{1+b^5c+bc^5} \le \frac{1}{3b^2c^2}=\frac{a^2}{3}$$ Hence

$$S \le \frac{a^2+b^2+c^2}{3}$$

Can we proceed with this?

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    $\begingroup$ What is the sum being taken over? $\endgroup$ – TomGrubb Oct 6 '18 at 3:37
  • $\begingroup$ Is the summation symbol representing a cyclic sum? If so, better write $\sum_{\mathrm{cyc}}$ than just $\sum$. $\endgroup$ – YiFan Oct 6 '18 at 5:45
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Because by Rearrangement $$b^4+c^4=b^3\cdot b+c^3\cdot c\geq b^3c+c^3b$$ and since $a=a^2bc$, we obtain: $$\sum_{cyc}\frac{a}{b^4+c^4+a}\leq\sum_{cyc}\frac{a}{b^3c+bc^3+a^2bc}=\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}=1.$$

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