Cho a, b, c, d > 0 và ab + bc + cd + da = 1. Chứng minh rằng:

\[\frac{{{a^3}}}{{b + c + d}} + \frac{{{b^3}}}{{c + d + a}} + \frac{{{c^3}}}{{a + b + c}} \ge \frac{1}{3}\]

Theo AM-GM ta có: \(\left\{ {\begin{array}{*{20}{c}}{\frac{{{a^3}}}{{b + c + d}} + \frac{{a\left( {b + c + d} \right)}}{9} \ge \frac{2}{3}{a^2}}\\{\frac{{{b^3}}}{{c + d + a}} + \frac{{b\left( {c + d + a} \right)}}{9} \ge \frac{2}{3}{b^2}}\\{\frac{{{c^3}}}{{d + a + b}} + \frac{{c\left( {d + a + b} \right)}}{9} \ge \frac{2}{3}{c^2}}\end{array}} \right.\)

\( \Rightarrow \frac{{{a^3}}}{{b + c + d}} + \frac{{{b^3}}}{{c + d + a}} + \frac{{{c^3}}}{{d + a + b}} + \frac{{{d^3}}}{{a + b + c}} + \frac{{2\left( {ab + ac + ad + bc + bd + cd} \right)}}{9}\)

\( \ge \frac{2}{3}\left( {{a^2} + {b^2} + {c^2} + {d^2}} \right)\,\,\,\left( 1 \right)\)

Theo AM-GM ta có:

\(3\left( {{a^2} + {b^2} + {c^2} + {d^2}} \right) = \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + {c^2}} \right) + \left( {{a^2} + {d^2}} \right) + \left( {{b^2} + {c^2}} \right) + \left( {{b^2} + {d^2}} \right) + \left( {{c^2} + {d^2}} \right)\)

\( \ge 2\left( {ab + ac + ad + bc + bd + cd} \right)\)

\( \Rightarrow \frac{1}{3}\left( {{a^2} + {b^2} + {c^2} + {d^2}} \right) \ge \frac{2}{9}\left( {ab + ac + ad + bc + bd + cd} \right)\,\,\,\left( 2 \right)\)

Từ (1) và (2) suy ra:

\(\frac{{{a^3}}}{{b + c + d}} + \frac{{{b^3}}}{{c + d + a}} + \frac{{{c^3}}}{{d + a + b}} + \frac{{{d^3}}}{{a + b + c}} \ge \frac{1}{3}\left( {{a^2} + {b^2} + {c^2} + {d^2}} \right)\,\,\,\left( 3 \right)\)

Mặt khác ta có:

\({a^2} + {b^2} + {c^2} + {d^2} = \frac{{{a^2} + {b^2}}}{2} + \frac{{{b^2} + {c^2}}}{2} + \frac{{{c^2} + {d^2}}}{2} + \frac{{{d^2} + {a^2}}}{2} \ge ab + bc + cd + da = 1\,\,\,\left( 4 \right)\)

Từ (3) và (4) suy ra:

\(\frac{{{a^3}}}{{b + c + d}} + \frac{{{b^3}}}{{c + d + a}} + \frac{{{c^3}}}{{d + a + b}} + \frac{{{d^3}}}{{a + b + c}} \ge \frac{1}{3}\).

Dấu "=" xảy ra khi: \(a = b = c = d = \frac{1}{2}\).