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

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