Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

Hint: By AM-GM, $$(abc+a) + (b+c) \geqslant 2a\sqrt{bc}+2\sqrt{bc} = 2\sqrt{bc}(1+a)$$

Now multiply three such inequalities (after cyclical shift).

After replacing $a\rightarrow\frac{1}{a},$ $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$ we need to prove that $$(1+ab+ac+bc)^3\geq8abc(1+a)(1+b)(1+c).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.

Thus, we need to prove that: $$(1+3v^2)^3\geq8w^3(1+w^3+3v^2+3u)$$ or $$(1+3v^2)\geq8(w^3+w^6+3v^3w^2+3uw^3)$$ and since by AM-GM $v\geq w$ and $v^4\geq uw^3,$ it's enough to prove that: $$(1+3v^2)^3\geq8(v^3+v^6+3v^5+3v^4)$$ or $$(1+3v^2)^3\geq8v^3(1+v)^3$$ or $$1+3v^2\geq2v(1+v)$$ or $$(1-v)^2\geq0$$ and we are done!

Also, $uvw$ kills this inequality, but it's not so nice.