Question

Cho x, y, z là các số thực dương thỏa mãn điều kiện xyz=8

Chứng minh rằng \(\frac{1}{2x+y+6}+\frac{1}{2y+z+6}+\frac{1}{2z+x+6}\le\frac{1}{4}\)

Answer 1

Đặt \(\left(x;y;z\right)=\left(2a^2;2b^2;2c^2\right)\Rightarrow abc=1\)

\(VT=\frac{1}{4a^2+2b^2+6}+\frac{1}{4b^2+2c^2+6}+\frac{1}{4c^2+2a^2+6}\)

\(VT=\frac{1}{\left(2a^2+2\right)+\left(2a^2+2b^2\right)+4}+\frac{1}{\left(2b^2+2\right)+\left(2b^2+2c^2\right)+4}+\frac{1}{\left(2c^2+2\right)+\left(2c^2+2a^2\right)+4}\)

\(VT\le\frac{1}{4a+4ab+4}+\frac{1}{4b+4bc+4}+\frac{1}{4c+4ca+4}=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\) hay \(x=y=z=2\)