TITLE: if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: By Holder $$\prod\limits_{cyc}(a+1)\geq\left(\sqrt[3]{abc}+1\right)^3=27$$