Question

chứng minh rằng : \(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-3\sqrt{2}}\right)^8\ge3^6\)

@Akai Haruma , @Lightning Farron

Answer 1

Hung nguyen : help

Answer 2

Đặt A = \(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(\Rightarrow\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-3\sqrt{2}}< A\)

\(A^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{9-8}=9\)

\(\Rightarrow A^8=\left(A^3\right)^2.A^2=9^2.\left(\sqrt[3]{9}\right)^2=3^4.\sqrt[3]{81}=3^5.\sqrt[3]{3}< 3^6\)

\(\Rightarrow\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-3\sqrt{2}}< A< 3^6\)

......... Kaito Kid ........