Áp dụng BĐT Schwartz ta có:
\(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{x+2y+z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\ge\frac{16}{x+y+2z}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{4.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{16}=1\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\frac{3}{4}\)