Question

[Phạm Thị Hằng]

Find the maximum value of \(M=\frac{x}{2x+y}+\frac{y}{2y+z}+\frac{z}{2z+x}\) , x,y,z > 0

Answer 1

[alibaba nguyễn]

Ta có

\(1-\frac{2x}{2x+y}=1-\frac{2xy}{2xy+y^2}=\frac{y^2}{2xy+y^2}\left(1\right)\)

Ta lại có

\(\frac{y^2}{2xy+y^2}+\frac{2xy+y^2}{\left(x+y+z\right)^2}\ge\frac{2y}{x+y+z}\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow1-\frac{2x}{2x+y}+\frac{2xy+y^2}{\left(x+y+z\right)^2}\ge\frac{2y}{x+y+z}\left(3\right)\)

Tương tự

\(1-\frac{2y}{2y+z}+\frac{2yz+z^2}{\left(x+y+z\right)^2}\ge\frac{2z}{\left(x+y+z\right)}\left(4\right)\)

\(1-\frac{2z}{2z+x}+\frac{2xz+x^2}{\left(x+y+z\right)^2}\ge\frac{2x}{x+y+z}\left(5\right)\)

Lấy (3) + (4) + (5) vế theo vế ta được

\(3-2M+\frac{2\left(xy+yz+zx\right)+x^2+y^2+z^2}{\left(x+y+z\right)^2}\ge\frac{2\left(x+y+z\right)}{x+y+z}\)

\(\Leftrightarrow3-2M+1\ge2\)

\(\Leftrightarrow M\le1\)

Dấu =  xảy ra khi \(x=y=z\)