if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
 A: Using $\bf{A.M\geq G.M}$
$$\frac{a}{2}+\frac{a}{2}+1\geq 3\left(\frac{a^2}{4}\right)^{\frac{1}{3}}$$
Similarly $$\frac{b}{2}+\frac{b}{2}+1\geq 3\left(\frac{b^2}{4}\right)^{\frac{1}{3}}$$
$$\frac{c}{2}+\frac{c}{2}+1\geq 3\left(\frac{c^2}{4}\right)^{\frac{1}{3}}$$
So $$(a+1)(b+1)(c+1)\geq 27 \left(\frac{(abc)^2}{64}\right)^{\frac{1}{3}} = 27$$
equality hold when $\displaystyle a= b=c = 2$
A: We have  $$(a+1)(b+1)(c+1)=9+a+b+c+ab+bc+ca,$$
since $$abc=8.$$ We get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}=2$$ and $$\frac{ab+bc+ca}{3}\geq \sqrt[3]{(abc)^2}=4$$ Multplying by $3$ and adding up, we get the statement above.
A: Here is a really hands-on way to proceed.
Suppose $b\gt c$, replace both with $B, C=\sqrt {bc}$ which keeps the product the same, but makes the two equal. Now note that $$0\lt (\sqrt b-\sqrt c)^2=b+c-2\sqrt {bc}$$ or $$B+C=2\sqrt{bc}\lt b+c$$ so the sum is reduced, and looking at the components of $(a+1)(b+1)(c+1)$ we have $B+C\lt b+c$ and $a(B+C)\lt a(b+c)$ with the other terms unchanged.
The replacement therefore reduces the product, which can be reduced, therefore, unless all the terms are equal. The minimum value therefore has $a=b=c$.
Note this includes a hidden proof of the AM/GM inequality for two variables - the AM/GM inequality is the essential idea here (or this can be seen as a property of curvature - the sum being flat and the product being curved). Any proof will also use somewhere the fact that the numbers are stated to be positive (else set one of them equal to $-1$ ...). You should reflect on how the various proofs use this fact.
A: By Holder $$\prod\limits_{cyc}(a+1)\geq\left(\sqrt[3]{abc}+1\right)^3=27$$
