Cho x, y, z > 0 và 13x + 5y + 12z = 9.

Tìm max của \(M = \frac{{xy}}{{2x + y}} + \frac{{3yz}}{{2y + z}} + \frac{{6zx}}{{2z + x}}\).

Áp dụng bất đẳng thức \(\frac{a}{{b + c + d}} \le \frac{1}{9}\left( {\frac{a}{b} + \frac{a}{c} + \frac{a}{d}} \right)\) ta có:

\(\frac{{xy}}{{2x + y}} \le \frac{1}{9}\left( {\frac{{xy}}{x} + \frac{{xy}}{x} + \frac{{xy}}{y}} \right) = \frac{1}{9}\left( {y + y + x} \right) = \frac{1}{9}(2y + x)\)

\(\frac{{3yz}}{{2y + z}} \le 3.\frac{1}{9}\left( {\frac{{yz}}{y} + \frac{{yz}}{y} + \frac{{yz}}{z}} \right) = \frac{1}{3}\left( {z + z + y} \right) = \frac{1}{3}(2z + y)\)

\(\frac{{6xz}}{{2z + x}} \le 6.\frac{1}{9}\left( {\frac{{xz}}{z} + \frac{{xz}}{z} + \frac{{xz}}{x}} \right) = \frac{2}{3}\left( {x + x + z} \right) = \frac{1}{9}(2x + z)\)

\(\begin{array}{l} \Rightarrow M = \frac{{xy}}{{2x + y}} + \frac{{3yz}}{{2y + z}} + \frac{{6zx}}{{2z + x}} \le \frac{1}{9}\left( {2y + z} \right) + \frac{1}{3}\left( {2z + y} \right) + \frac{2}{3}\left( {2x + z} \right)\\ \Rightarrow M = \frac{{xy}}{{2x + y}} + \frac{{3yz}}{{2y + z}} + \frac{{6zx}}{{2z + x}} \le \frac{{13}}{9}x + \frac{5}{9}y + \frac{{12}}{9}z = \frac{1}{9}\left( {13x + 5y + 12z} \right) = \frac{1}{9}.9 = 1\end{array}\)

Dấu “=” xảy ra ⇔ x = y = z = \(\frac{3}{{10}}\).