Question

Cho a,b,c >0 t/m a+b+c=3

Tìm GTNN của \(P=\left(1+\frac{3}{a}\right)\left(1+\frac{3}{b}\right)\left(1+\frac{3}{c}\right)\)

Answer 1

a+b+c=3

Mà: \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le1\Rightarrow abc\le1\Rightarrow\frac{1}{abc}\ge1\)(cô-si)

\(P=1+\frac{3}{a}+\frac{3}{b}+\frac{3}{c}+\frac{9}{ab}+\frac{9}{ac}+\frac{9}{bc}+\frac{27}{abc}\)

\(=1+3.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+\frac{27}{abc}\)

\(\ge1+9\sqrt[3]{\frac{1}{abc}}+27\sqrt[3]{\frac{1}{\left(abc\right)^2}}+\frac{27}{abc}\ge1+9\sqrt[3]{\frac{1}{1}}+27\sqrt[3]{\frac{1}{1}}+\frac{27}{1}=64\)

Vậy GTNN của P là 64 tại a=b=c=1