Question

Cho a, b, c > 0 thỏa mãn: a + b + c = 2

CM: \(\sqrt[3]{\frac{1}{a}}+\sqrt[3]{\frac{1}{b}}+\sqrt[3]{\frac{1}{c}}\ge\frac{3\sqrt[3]{12}}{2}\)

Answer 1

Đặt \(\left(a;b;c\right)=\left(x^3;y^3;z^3\right)\Rightarrow x^3+y^3+z^3=2\)

Ta có: \(x^3+\frac{2}{3}+\frac{2}{3}\ge3\sqrt[3]{x^3.\frac{4}{9}}=\sqrt[3]{12}x\)

Tương tự: \(y^3+\frac{2}{3}+\frac{2}{3}\ge\sqrt[3]{12}y\); \(z^3+\frac{2}{3}+\frac{2}{3}\ge\sqrt[3]{12}z\)

\(\Rightarrow x^3+y^3+z^3+4\ge\sqrt[3]{12}\left(x+y+z\right)\)

\(\Rightarrow x+y+z\le\frac{6}{\sqrt[3]{12}}=\sqrt[3]{18}\)

Ta có: \(P=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\ge\frac{9}{\sqrt[3]{18}}=\frac{3\sqrt[3]{12}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt[3]{\frac{2}{3}}\) hay \(a=b=c=\frac{2}{3}\)