\(A=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(\Leftrightarrow A=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\)
\(\Leftrightarrow A=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\left(1\right)\)
Theo BĐT Cô si, ta có :
\(\left[\left(x+1\right)+\left(y+1\right)+\left(z+1\right)\right]\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\ge9\)
\(\Leftrightarrow4\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\ge9\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{9}{4}\left(3\right)\)
\(\Leftrightarrow A=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\ge\dfrac{9}{4}\)
\(\Leftrightarrow A=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}.\)