# Prove $\sum \frac{a}{a+b^4+c^4} \le 1$

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?

• What is the sum being taken over? – TomGrubb Oct 6 '18 at 3:37
• Is the summation symbol representing a cyclic sum? If so, better write $\sum_{\mathrm{cyc}}$ than just $\sum$. – YiFan Oct 6 '18 at 5:45

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: $$\sum_{cyc}\frac{a}{b^4+c^4+a}\leq\sum_{cyc}\frac{a}{b^3c+bc^3+a^2bc}=\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}=1.$$