\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\le\frac{1}{16}\left(\frac{x}{x}+\frac{x}{y}+\frac{x}{x}+\frac{x}{z}\right)=\frac{1}{16}\left(2+\frac{x}{y}+\frac{x}{z}\right)\)
\(tươngtự:\frac{y}{2y+z+x}\le\frac{1}{16}\left(2+\frac{y}{z}+\frac{y}{x}\right);\frac{z}{2z+x+y}\le\frac{1}{16}\left(2+\frac{z}{x}+\frac{z}{y}\right).\text{Cộng vế theo vế ta được:}\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+y+x}\le\frac{1}{16}\left(2+2+2+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}+\frac{z}{y}\right)=\frac{1}{16}\left[6+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\right]\ge\frac{1}{16}\left(6+2\sqrt{\frac{xy}{xy}}+2\sqrt{\frac{xz}{xz}}+2\sqrt{\frac{yz}{yz}}\right)=\)
\(=\frac{12}{16}=\frac{3}{4}\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{3}{4}\left(\text{đpcm}\right)\)
