Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
Su\,dung\,\frac{1}{x} + \frac{1}{y} \ge \frac{4}{{x + y}}\\
Ta\,co:\frac{1}{{2x + y + z}} = \frac{1}{4}.\frac{4}{{\left( {x + y} \right) + \left( {x + z} \right)}}\\
\le \frac{1}{4}\left( {\frac{1}{{x + y}} + \frac{1}{{x + z}}} \right) = \frac{1}{{16}}\left( {\frac{4}{{x + y}} + \frac{4}{{x + z}}} \right)\\
\le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{x} + \frac{1}{z}} \right) = \frac{1}{{16}}\left( {\frac{2}{x} + \frac{1}{y} + \frac{1}{z}} \right)\\
Tuong\,tu\,\frac{1}{{x + 2y + z}} \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{2}{y} + \frac{1}{z}} \right)\\
\frac{1}{{x + y + 2z}} \le \frac{1}{{16}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{2}{z}} \right)\\
\Rightarrow \frac{1}{{2x + y + z}} + \frac{1}{{x + 2y + z}} + \frac{1}{{x + y + 2z}} \le \frac{1}{{16}}\left( {\frac{4}{x} + \frac{4}{y} + \frac{4}{z}} \right)\\
= \frac{1}{4}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = \frac{1}{4}.8 = 2\\
Dau\, = \,xay\,ra\,khi\,x = y = z = \frac{3}{8}
\end{array}$
