Áp dụng bất đẳng thức $B-C-S$ dạng Engel như sau:
$\dfrac{1}{x} + \dfrac{1}{y} \ge \dfrac{4}{{x + y}}$
Áp dụng vào bất đẳng thức dưới ta được:
$\begin{array}{l}
(B - C - S)\\
\dfrac{1}{{2a + b + c}} = \dfrac{1}{{\left( {a + b} \right) + \left( {a + c} \right)}} \le \dfrac{1}{4}\left( {\dfrac{1}{{a + b}} + \dfrac{1}{{a + c}}} \right)\\
\dfrac{1}{{2b + a + c}} = \dfrac{1}{{\left( {a + b} \right) + \left( {b + c} \right)}} \le \dfrac{1}{4}\left( {\dfrac{1}{{a + b}} + \dfrac{1}{{b + c}}} \right)\\
\dfrac{1}{{2c + a + b}} = \dfrac{1}{{\left( {a + c} \right) + \left( {c + b} \right)}} \le \dfrac{1}{4}\left( {\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}}} \right)\\
\Rightarrow M \le \dfrac{1}{4}\left( {\dfrac{2}{{a + b}} + \dfrac{2}{{b + c}} + \dfrac{2}{{c + a}}} \right)\\
\Rightarrow M \le \dfrac{1}{2}\left( {\dfrac{1}{{a + b}} + \dfrac{1}{{b + c}} + \dfrac{1}{{c + a}}} \right)\\
\Rightarrow M \le \dfrac{1}{2}.\left[ {\dfrac{1}{4}\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{c} + \dfrac{1}{a}} \right)} \right]\\
\Rightarrow M \le \dfrac{1}{2}\left[ {\dfrac{1}{4}.2.\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)} \right] = \dfrac{1}{8}.2.\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
= 2.\dfrac{1}{8}.4 = 1\\
\Rightarrow \max M = 1\\
' = ' \Leftrightarrow a = b = c = \dfrac{3}{4}
\end{array}$
