bổ sung đề: xyz=1
AD BĐT Cô si cho 3 số ta có:
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+x}\ge3\sqrt[3]{\dfrac{x^2}{1+y}.\dfrac{y^2}{1+z}.\dfrac{z^2}{1+x}}=3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Vì xyz=1 nên suy ra \(x\le1\Leftrightarrow x+1\le2\Leftrightarrow\dfrac{1}{x+1}\ge\dfrac{1}{2}\)
Tương tự ta có: \(\dfrac{1}{y+1}\ge\dfrac{1}{2}\)
\(\dfrac{1}{z+1}\ge\dfrac{1}{2}\)
Suy ra \(3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\ge3\sqrt[3]{\dfrac{1}{2.2.2}}=3\sqrt[3]{\dfrac{1}{8}}=\dfrac{3}{2}\)
Vậy \(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+x}\ge\dfrac{3}{2}khixyz=1\)
Có thiếu dữ kiện không vậy bạn ?
