\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).64}}=\dfrac{3x}{4}\)
\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)
\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{x+y+z}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{xyz}}{2}-\dfrac{3}{4}=\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)
(bài này chắc thiếu đk xyz=1 ?nên mình bổ sung xyz=1)
( xyz=3)
Áp dụng BDDT AM-GM:
Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).8.8}}=3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)
Chứng minh tương tự ta có:
\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)
\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)
Cộng từng vế ta được:
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{3+x+y+z}{4}\ge\dfrac{3\left(x+y+z\right)}{4}\)
\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{3x+3y+3z-3-x-y-z}{4}=\dfrac{2\left(x+y+z\right)-3}{4}\)
\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{2.\sqrt[3]{xyz}-3}{4}=\dfrac{2.3-3}{4}=\dfrac{3}{4}\left(đfcm\right)\)
