câu1 Tìm giá trị nhỏ nhất: y= ( a/3b+1 )( b/3c+1 )( c/3a+1 )≥ 8

Giải thích các bước giải:

$\dfrac{a}{3b}+1=\dfrac{a}{3b}+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\ge 4\sqrt[4]{\dfrac{a}{3b}\dfrac{1}{3}\dfrac{1}{3}\dfrac{1}{3}}=\dfrac{4}{3}.\sqrt[4]{\dfrac{a}{b}}$

$\dfrac{b}{3c}+1=\dfrac{b}{3c}+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\ge 4\sqrt[4]{\dfrac{b}{3c}\dfrac{1}{3}\dfrac{1}{3}\dfrac{1}{3}}=\dfrac{4}{3}.\sqrt[4]{\dfrac{b}{c}}$

$\dfrac{c}{3a}+1=\dfrac{c}{3a}+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\ge 4\sqrt[4]{\dfrac{c}{3a}\dfrac{1}{3}\dfrac{1}{3}\dfrac{1}{3}}=\dfrac{4}{3}.\sqrt[4]{\dfrac{c}{a}}$

$\rightarrow (\dfrac{a}{3b}+1)(\dfrac{b}{3c}+1)(\dfrac{c}{3a}+1)\ge \dfrac{4}{3}.\sqrt[4]{\dfrac{a}{b}}. \dfrac{4}{3}.\sqrt[4]{\dfrac{b}{c}}.\dfrac{4}{3}.\sqrt[4]{\dfrac{c}{a}}=(\dfrac{4}{3})^3$