Question

Cho x, y, z > 0 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

Tìm GTLN của \(P=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)

Giúp.

Answer 1

Ta có:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};\frac{1}{y}+\frac{1}{x}\ge\frac{4}{x+y}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\ge\frac{4}{x+y}+\frac{4}{x+y}\ge\frac{16}{x+2y+z}\Rightarrow\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)\(TT:\)

\(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right);\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)\\(S\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=1\)