Question

Tìm GTNN của \(P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\) trong đó các số dương a,b,c thỏa mãn: \(a+b+c\le\frac{3}{2}\)

Answer 1

\(P\ge\left(3+\frac{4}{a+b}\right)\left(3+\frac{4}{b+c}\right)\left(3+\frac{4}{c+a}\right)\)

Đặt \(\left(\frac{4}{a+b};\frac{4}{b+c};\frac{4}{c+a}\right)=\left(x;y;z\right)\Rightarrow\frac{3}{4}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\)

\(\Rightarrow xyz\ge64\)

\(P=\left(3+x\right)\left(3+y\right)\left(3+z\right)\)

\(P=xyz+3\left(xy+yz+zx\right)+9\left(x+y+z\right)+27\)

\(P\ge xyz+3.3\sqrt[3]{\left(xyz\right)^2}+9.3\sqrt[3]{xyz}+27\)

\(P\ge64+3.3.\sqrt[3]{64^2}+27.4+27=...\)