To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M  G.M  inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals
$$4abc-c^3 < 4abc$$ so
$$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$  but i am unable to proceed here

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 $$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\geq(a+b)^2\implies \frac{a^2}{b}+\frac{b^2}{a}\geq a+b$$
