\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=1-\frac{1}{x+1}+1-\frac{1}{y+1}+1-\frac{1}{z+1}\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy-Schwraz dạng Engel ta có:
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}\)
\(\le3-\frac{3^2}{1+3}=3-\frac{9}{4}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)