Chứng minh 1/ (1 + a^3) + 1/ (1 + b^3) + 1/ (1 + c^3) > = 3 / (1 + abc)
Câu hỏi:
Chứng minh \(\frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} \ge \frac{3}{{1 + abc}}\) với a, b, c ≥ 1.
Trả lời:
Ta có:
\(\frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} \ge \frac{3}{{1 + abc}}\)
\( \Leftrightarrow \frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{4}{{1 + abc}}\)
Xét \(\frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {y^2}}} \ge \frac{2}{{1 + xy}}\) với x, y ≥ 1
\( \Leftrightarrow \frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {y^2}}} - \frac{2}{{1 + xy}} \ge 0\)
\( \Leftrightarrow \frac{{\left( {1 + {y^2}} \right)\left( {1 + xy} \right) + \left( {1 + {x^2}} \right)\left( {1 + xy} \right) - 2\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{1 + xy + {y^2} + x{y^3} + 1 + xy + {x^2} + {x^3}y - 2 - 2{{\rm{x}}^2} - 2{y^2} - 2{x^2}{y^2}}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{x{y^3} + 2xy + {x^3}y - {{\rm{x}}^2} - {y^2} - 2{x^2}{y^2}}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{xy\left( {{y^2} - 2{\rm{x}}y + {x^2}} \right) - \left( {{y^2} - 2{\rm{x}}y + {x^2}} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\)
\( \Leftrightarrow \frac{{{{\left( {x - y} \right)}^2}\left( {xy - 1} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\)
Vì x, y ≥ 1 nên xy – 1 ≥ 0
Mà (x – y)2 ≥ 0, 1 + x2 > 0, 1 + y2 > 0, xy + 1 > 0
Suy ra \(\frac{{{{\left( {x - y} \right)}^2}\left( {xy - 1} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\left( {1 + xy} \right)}} \ge 0\) với mọi x, y ≥ 1
Do đó \(\frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {y^2}}} \ge \frac{2}{{1 + xy}}\) với x, y ≥ 1
Áp dụng bất đẳng thức trên ta có:
\(\frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{2}{{1 + \sqrt {{a^3}{b^3}} }} + \frac{2}{{1 + \sqrt {ab{c^4}} }}\)
\( \Leftrightarrow \frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{4}{{1 + \sqrt {{a^3}{b^3}\sqrt {ab{c^4}} } }}\)
\( \Leftrightarrow \frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} + \frac{1}{{1 + abc}} \ge \frac{4}{{1 + abc}}\)
Dấu “ = ” xảy ra khi a = b = c = 1
Vậy \(\frac{1}{{1 + {a^3}}} + \frac{1}{{1 + {b^3}}} + \frac{1}{{1 + {c^3}}} \ge \frac{3}{{1 + abc}}\) với a, b, c ≥ 1.
