Question

Tìm GTLN của bt: P = \(\dfrac{x}{x+1}\)+ \(\dfrac{y}{y+1}+\dfrac{z}{z+1}\) với x,y,z > 0 và x+y+z = 1

Answer 1

Ta có: \(\dfrac{x}{x+1}=1-\dfrac{1}{x+1}\)

\(\dfrac{y}{y+1}=1-\dfrac{1}{y+1}\)

\(\dfrac{z}{z+1}=1-\dfrac{1}{z+1}\)

Cộng vế theo vế:

\(P=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\)

Áp dụng BĐT Cauchy- Schwarz dạng Engel:

\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z+3}=\dfrac{9}{4}\)

\(\Rightarrow P\le3-\dfrac{9}{4}=\dfrac{3}{4}\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)