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?

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.
