Câu hỏi của Tăng Quỳnh Chi

Question

Cho a,b,c > 0. Tìm GTNN của biểu thức:

$\left(1+\frac{a}{3b}\right)\left(1+\frac{b}{3c}\right)\left(1+\frac{c}{3a}\right)$

Nguyễn Việt Lâm · Accepted Answer

$P=\left(1+\frac{a}{3b}\right)\left(1+\frac{c}{3a}+\frac{b}{3c}+\frac{b}{9a}\right)$

$P=1+\frac{1}{3}\left(\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\right)+\frac{1}{9}\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\frac{1}{27}$

$P\ge1+\frac{1}{27}+\frac{1}{3}.3\sqrt[3]{\frac{abc}{abc}}+\frac{1}{9}.3\sqrt[3]{\frac{abc}{abc}}=\frac{64}{27}$

$\Rightarrow P_{min}=\frac{64}{27}$ khi $a=b=c$Câu trả lời: $P=\left(1+\frac{a}{3b}\right)\left(1+\frac{c}{3a}+\frac{b}{3c}+\frac{b}{9a}\right)$

$P=1+\frac{1}{3}\left(\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\right)+\frac{1}{9}\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\frac{1}{27}$

$P\ge1+\frac{1}{27}+\frac{1}{3}.3\sqrt[3]{\frac{abc}{abc}}+\frac{1}{9}.3\sqrt[3]{\frac{abc}{abc}}=\frac{64}{27}$

$\Rightarrow P_{min}=\frac{64}{27}$ khi $a=b=c$

Violympic toán 9