Question

Cho  \(x,y,z>0\) thỏa mãn \(x+y+z=1\). Tìm GTLN của  \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

Answer 1

\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

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

Áp dụng bddt Bunhiacopski dạng phân thức:

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

\(\Rightarrow-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le\frac{-9}{4}\)

\(\Rightarrow3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le\frac{3}{4}\)

Dấu "=" khi x = y = z = \(\frac{1}{3}\)

Answer 2

\(\frac{3}{4}-P=\Sigma\frac{x\left(y-z\right)^2}{4\left(x+1\right)\left(x+y\right)\left(x+z\right)}\ge0\)

Answer 3

Cách hack điểm hỏi đáp trên OLM: https://www.youtube.com/watch?v=sMvl8_N_N54