Question

[Tobot Z]

Cho 3 số thực dương x,y,z thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\).Tìm GTLN của biểu thức

\(P=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)

Answer 1

[Nguyễn Việt Lâm]

\(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=2\)

Lại có \(\dfrac{1}{2x+y+z}=\dfrac{1}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)

Tương tự \(\dfrac{1}{x+2y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\)

\(\dfrac{1}{x+y+2z}\le\dfrac{1}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)

Cộng vế với vế: \(P\le\dfrac{1}{2}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{1}{2}.2=1\)

\(\Rightarrow P_{max}=1\) khi \(x=y=z=\dfrac{3}{4}\)