Nếu x, y, z > 0 thỏa mãn: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4\) thì

\(\frac{1}{{2x + y + z}} + \frac{1}{{x + 2y + z}} + \frac{1}{{x + y + 2z}} \le 1\).

Ta có: x, y, z > 0

Áp dụng BĐT Cô si ta có:

\(\left( {x + y} \right) \ge 2\sqrt {xy} \) và \(\left( {\frac{1}{x} + \frac{1}{y}} \right) \ge 2\sqrt {\frac{1}{{xy}}} \)

\( \Rightarrow \left( {x + y} \right)\left( {\frac{1}{x} + \frac{1}{y}} \right) \ge 2\sqrt {xy} .2\sqrt {\frac{1}{{xy}}} = 4\)

\( \Leftrightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}} \Leftrightarrow \frac{1}{{x + y}} \le 4\left( {\frac{1}{x} + \frac{1}{y}} \right)\) (*)

Áp dụng (*) ta có:

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

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

\(\frac{1}{{x + y + 2z}} = \frac{1}{{x + z + y + z}} = \frac{1}{{\left( {x + z} \right) + \left( {y + z} \right)}} \le \frac{1}{4}\left( {\frac{1}{{x + z}} + \frac{1}{{y + z}}} \right) \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{z} + \frac{1}{y} + \frac{1}{z}} \right)\left( 3 \right)\)

Cộng hai vế của (1) , (2), (3) ta có:

\(\frac{1}{{2x + y + z}} + \frac{1}{{x + 2y + z}} + \frac{1}{{x + y + 2z}} \le 1\) (đcpcm).