\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=2\)
\(\Leftrightarrow\dfrac{1}{x+1}=1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}=\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)
Làm tương tự : \(\dfrac{1}{y+1}=1-\dfrac{1}{x+1}+1-\dfrac{1}{z+1}=\dfrac{x}{x+1}+\dfrac{z}{z+1}\ge2\sqrt{\dfrac{xz}{\left(x+1\right)\left(z+1\right)}}\)
\(\dfrac{1}{z+1}=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}=\dfrac{x}{x+1}+\dfrac{y}{y+1}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)
\(\Rightarrow\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}\ge8\dfrac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow1\ge8xyz\)
\(\Leftrightarrow xyz\le\dfrac{1}{8}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)
P/s : Bạn chịu khó tìm câu hỏi tương tự trc khi hỏi nhé.
