To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$

Notice that $$\sum_{cyc}\frac{abc}{a^3+b^3+abc} = \sum_{cyc}\frac{1}{\frac{a^2}{bc}+\frac{b^2}{ac}+1} \leq \sum_{cyc}\frac{1}{\frac{a+b}{c}+1} = \sum_{cyc}\frac{c}{a+b+c} = 1 $$ Then divide both sides by $abc$.

Note: In the inequality part we used Cauchy-Schwartz to get $$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\geq(a+b)^2\implies \frac{a^2}{b}+\frac{b^2}{a}\geq a+b$$

$$\sum_{cyc}\frac{abc}{a^3+b^3+abc}\leq\sum_{cyc}\frac{abc}{a^2b+ab^2+abc}=\sum_{cyc}\frac{c}{a+b+c}=1$$ because by Rearrangement $a^3+b^3=a^2\cdot a+b^2\cdot b\geq a^2b+b^2a$.