Given three positive numbers $a,\,b,\,c$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqslant 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!).$$
My own problem is given a solution, and I'm looking forward to seeing a nicer one(s), thank you !
Solution. Without loss of generality, we can suppose $(a- 1)(b- 1)\geqslant 0 \rightarrow 1+ ab\geqslant a+ b$. Then
\begin{align*} (abc+ a+ b+ c)^{3}&= \left ( c(1+ ab)+ a+ b \right )^{2}\left ( c(1+ ab)+ a+ b \right )\\& \geqslant 4\,c(1+ ab)(a+ b)\left ( c(a+ b)+ (a+ b) \right )\\&= 2c(a+ b)^{2}(2+ 2ab)(1+ c)\\& \end{align*}
Again, by using a.m.-g.m.-inequality, we have $$2c\left(a+ b\right)^{2}\left( 2+ 2ab\right)\left( 1+ c\right)\geqslant 8abc\left(1+ a\right)\left(1+ b\right)\left(1+ c\right)$$
That is q.e.d!
Best Answer
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).