Cho các số thực dương a, b, c. CMR: \left( {{a^2} + 2} \right)\left( {{b^2} + 2} \right)\left( {{c^2} + 2} \right) \ge 9\left( {ab + bc + ca} \right)\left( * \right).
Giải:\begin{array}{l} V{T_{\left( * \right)}} = {\left( {abc} \right)^2} + 2\left[ {{{\left( {ab} \right)}^2} + {{\left( {bc} \right)}^2} + {{\left( {ca} \right)}^2}} \right] + 4\left( {{a^2} + {b^2} + {c^2}} \right) + 8\\ = \left[ {{{\left( {abc} \right)}^2} + 1 + 1} \right] + 2\left[ {{{\left( {ab} \right)}^2} + {{\left( {bc} \right)}^2} + {{\left( {ca} \right)}^2} + 3} \right] + 4\left( {{a^2} + {b^2} + {c^2}} \right)\, \end{array}
Theo bất đẳng thức Schur ta thấy:\begin{array}{l} \left\{ \begin{array}{l} 0 < \forall \,a,b,c \in R\\ 0 < \forall \,k \in R \end{array} \right. \Rightarrow {a^k}\left( {a - b} \right)\left( {a - c} \right) + {b^k}\left( {b - c} \right)\left( {b - a} \right) + {c^k}\left( {c - a} \right)\left( {c - b} \right) \ge 0\\ Khi\,k = 1:\,a\left( {a - b} \right)\left( {a - c} \right) + b\left( {b - c} \right)\left( {b - a} \right) + c\left( {c - a} \right)\left( {c - b} \right) \ge 0\left( i \right)\\ Coi\,p = a + b + c;\,q = ab + bc + ca\& r = abc \Rightarrow \left( i \right) \Leftrightarrow r \ge \frac{{p\left( {4q - {p^2}} \right)}}{9} \Leftrightarrow \frac{{9r}}{p} \ge 4q - {p^2}\left( {ii} \right) \end{array}
Áp dụng bất đẳng thức (ii) ta có:\begin{array}{l}
Do\,{\left( {abc} \right)^2} + 1 + 1 \ge 3\sqrt[3]{{{{\left( {abc} \right)}^2}}}\left( {Cauchy} \right) = \frac{{9abc}}{{3\sqrt[3]{{abc}}}} \ge \frac{{9abc}}{{a + b + c}}\left( {Cauchy} \right) \ge 4\left( {ab + bc + ca} \right) - {\left( {a + b + c} \right)^2}\\
Do\,\left\{ \begin{array}{l}
{\left( {ab} \right)^2} + 1 \ge 2ab\left( {Cauchy} \right)\\
{\left( {bc} \right)^2} + 1 \ge 2bc\left( {Cauchy} \right)\\
{\left( {ca} \right)^2} + 1 \ge 2ca\left( {Cauchy} \right)
\end{array} \right. \Rightarrow 2\left[ {{{\left( {ab} \right)}^2} + {{\left( {bc} \right)}^2} + {{\left( {ca} \right)}^2}} \right] + 6 \ge 4\left( {ab + bc + ca} \right)\\
Do\,\left\{ \begin{array}{l}
{a^2} + {b^2} \ge 2ab\left( {Cauchy} \right)\\
{b^2} + {c^2} \ge 2bc\left( {Cauchy} \right)\\
{c^2} + {a^2} \ge 2ca\left( {Cauchy} \right)
\end{array} \right. \Rightarrow 4\left( {{a^2} + {b^2} + {c^2}} \right) \ge 4\left( {ab + bc + ca} \right)\\
\Rightarrow V{T_{\left( * \right)}} \ge 4\left( {ab + bc + ca} \right) - {\left( {a + b + c} \right)^2} + 4\left( {ab + bc + ca} \right) + 4\left( {{a^2} + {b^2} + {c^2}} \right) = 8\left( {ab + bc + ca} \right)\\
- \left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + bc + ca} \right) + 4\left( {{a^2} + {b^2} + {c^2}} \right) = 6\left( {ab + bc + ca} \right) + 3\left( {{a^2} + {b^2} + {c^2}} \right)\\
Do\,\left\{ \begin{array}{l}
{a^2} + {b^2} \ge 2ab\left( {Cauchy} \right)\\
{b^2} + {c^2} \ge 2bc\left( {Cauchy} \right)\\
{c^2} + {a^2} \ge 2ca\left( {Cauchy} \right)
\end{array} \right. \Rightarrow \left( {{a^2} + {b^2} + {c^2}} \right) \ge \left( {ab + bc + ca} \right) \Rightarrow V{T_{\left( * \right)}} \ge 9\left( {ab + bc + ca} \right) = V{P_{\left( * \right)}}
\end{array}
Vậy BĐT được chứng minh. Dấu "=" xảy ra khi và chỉ khi a=b=c.
Không có nhận xét nào:
Đăng nhận xét