Question

cho x,y,z là các số dương t/m x+y+z=6

cm : \(8^x+8^y+8^z\ge4^{x+1}+4^{y+1}+4^{z+1}\)

please help me

Answer 1

Dự đoán dấu bằng xảy ra khi \(x=y=z=2\), áp dụng BĐT AM-GM ta có:

\(8^x+8^x+64\ge3\sqrt[3]{8^x\cdot8^x\cdot64}=12\cdot4^x\)

\(8^y+8^y+64\ge3\sqrt[3]{8^y\cdot8^y\cdot64}=12\cdot4^y\)

\(8^z+8^z+64\ge3\sqrt[3]{8^z\cdot8^z\cdot64}=12\cdot4^z\)

Suy ra \(2\left(8^x+8^y+8^z\right)+3\cdot64\ge12\left(4^x+4^y+4^z\right)\left(1\right)\)

Theo giả thiết ta có:

\(8^x+8^y+8^z\ge3\sqrt[3]{8^{x+y+z}}=3\sqrt[3]{8^6}=3\cdot64\left(2\right)\)

Cộng (1) với (2) theo vế ta có:

\(3\left(8^x+8^y+8^z\right)\ge12\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)