Giải thích các bước giải:
Áp dụng bất đẳng thức \(\frac{1}{a} + \frac{1}{b} \ge \frac{4}{{a + b}}\) ta có:
\(\begin{array}{l}
\frac{1}{{2a + b + c}} + \frac{1}{{a + 2b + c}} + \frac{1}{{a + b + 2c}}\\
= \frac{1}{4}\left( {\frac{4}{{\left( {a + b} \right) + \left( {a + c} \right)}} + \frac{4}{{\left( {a + b} \right) + \left( {b + c} \right)}} + \frac{4}{{\left( {a + c} \right) + \left( {b + c} \right)}}} \right)\\
\le \frac{1}{4}\left( {\frac{1}{{a + b}} + \frac{1}{{a + c}} + \frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{a + c}} + \frac{1}{{b + c}}} \right)\\
\le \frac{1}{4}.\frac{1}{4}\left( {\frac{4}{{a + b}} + \frac{4}{{a + c}} + \frac{4}{{a + b}} + \frac{4}{{b + c}} + \frac{4}{{a + c}} + \frac{4}{{b + c}}} \right)\\
\le \frac{1}{{16}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{a} + \frac{1}{c} + \frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{c} + \frac{1}{b} + \frac{1}{c}} \right)\\
\le \frac{1}{{16}}\left( {\frac{4}{a} + \frac{4}{b} + \frac{4}{c}} \right) = \frac{1}{{4a}} + \frac{1}{{4b}} + \frac{1}{{4c}}
\end{array}\)
