Chứng minh với a, b, c > = 1 ta có: 1 / (a^3 + 1) + 1 / (b^3 + 1) + 1 / (c^3 + 1)
Chứng minh với a, b, c ≥ 1 ta có: \[\frac{1}{{{a^3} + 1}} + \frac{1}{{{b^3} + 1}} + \frac{1}{{{c^3} + 1}} \ge \frac{3}{{1 + abc}}\].
Ta cần chứng minh BĐT phụ:
\(\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} \ge \frac{2}{{1 + ab}}\) với a, b > 0, ab ≥ 1
\( \Leftrightarrow \left( {\frac{1}{{1 + {a^2}}} - \frac{1}{{1 + ab}}} \right) + \left( {\frac{1}{{1 + {b^2}}} - \frac{1}{{1 + ab}}} \right) \ge 0\)
\( \Leftrightarrow \frac{{ab - {a^2}}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)}} + \frac{{ab - {b^2}}}{{\left( {1 + {b^2}} \right)\left( {1 + ab} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{\left( {ab - {a^2}} \right)\left( {1 + {b^2}} \right) + \left( {ab - {b^2}} \right)\left( {1 + {a^2}} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {ab - 1} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)}} \ge 0\) luôn đúng với \(\forall \)a,b > 0, ab ≥ 1
Ta có: \[\frac{1}{{{a^3} + 1}} + \frac{1}{{{b^3} + 1}} + \frac{1}{{{c^3} + 1}} \ge \frac{3}{{1 + abc}}\]
\( \Leftrightarrow \frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{4}{{1 + abc}}\)
Áp dụng ĐBT trên ta có:
\(VT = \frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{2}{{1 + \sqrt {{a^3}{b^3}} }} + \frac{2}{{1 + \sqrt {ab{c^4}} }}\)
\(VT \ge \frac{4}{{1 + \sqrt {\sqrt {{a^3}{b^3}.ab{c^4}} } }} = \frac{4}{{1 + abc}}\)
Vậy \[\frac{1}{{{a^3} + 1}} + \frac{1}{{{b^3} + 1}} + \frac{1}{{{c^3} + 1}} \ge \frac{3}{{1 + abc}}\] (đpcm).