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

Question

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!

Answer 1

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.

Answer 2

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).