Question

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

tìm GTLN của: \(M=\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\)

Answer 1

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(\Leftrightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\le1\)

\(\frac{1}{x+y}+\frac{1}{y+z}\ge\frac{4}{2x+y+z}\Rightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge4\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)

\(4M\le1\Leftrightarrow M\le\frac{1}{4}\)     \(M=\frac{1}{4}\Leftrightarrow x=y=z=3\)