<s>[INST] 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
 [/INST] The best proof is use $a^3+b^3 \geqslant ab(a+b).$ I have another proof
Because
$$(2c^2+ab)(a^3+abc+b^3)-abc\left[2(a^2+b^2+c^2)+ab+bc+ca\right]$$
$$=\frac{c(ab+bc-2ca)^2}2+\frac{c(3a^2+4ab+2ac+3c^2)(a-b)^2+b(a^2+5b^2)(c-a)^2}{4} \geqslant 0.$$
Thefore
$$\frac{abc}{a^3+abc+b^3} \leqslant \frac{2c^2+ab}{2(a^2+b^2+c^2)+ab+bc+ca},$$
so
$$\sum \frac{abc}{a^3+abc+b^3} \leqslant \sum \frac{2c^2+ab}{2(a^2+b^2+c^2)+ab+bc+ca}=1.$$
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