Cho phương trình : \(a{x^3} + b{x^2} + cx + d = 0\...

- Hướng dẫn giải

Phương pháp giải:

Sử dụng BĐT Bunhiacopxi cho 3 số.

Giải chi tiết:

Phương trình : \(a{x^3} + b{x^2} + cx + d = 0\,\,\,\left( {a \ne 0} \right)\) có 3 nghiệm dương \({x_1};{x_2};{x_3}\)

 \(\begin{array}{l} \Rightarrow \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right) = 0\\ \Leftrightarrow {x^3} - \left( {{x_1} + {x_2} + {x_3}} \right){x^2} + \left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)x - {x_1}{x_2}{x_3} = 0\,\,\,\left( 1 \right)\end{array}\)

Mà \(a{x^3} + b{x^2} + cx + d = 0\,\,\,\left( {a \ne 0} \right) \Leftrightarrow {x^3} + \frac{b}{a}{x^2} + \frac{c}{a}x + \frac{d}{a} = 0\,\,\,\left( 2 \right)\)

Từ (1) và (2) \( \Rightarrow \left\{ \begin{array}{l}{x_1} + {x_2} + {x_3} = \frac{{ - b}}{a}\\{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2} = \frac{c}{a}\\{x_1}{x_2}{x_3} = \frac{{ - d}}{a}\end{array} \right.\)

Bài toán \(x_1^7 + x_2^7 + x_3^7 \ge \frac{{ - {b^3}{c^2}}}{{81{a^5}}}\) trở thành

Chưng minh: \(x_1^7 + x_2^7 + x_3^7 \ge \frac{1}{{81}}{\left( {{x_1} + {x_2} + {x_3}} \right)^3}{\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)^2}\,\,\,\left( * \right)\)

Thật vậy ta có : với \({x_1};{x_2};{x_3}\) là các số dương. Áp dụng BĐT Cô-si 2 số ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}x_1^2 + x_2^2 \ge 2{x_1}{x_2}\\x_3^2 + x_2^2 \ge 2{x_3}{x_2}\\x_1^2 + x_3^2 \ge 2{x_1}{x_3}\end{array} \right. \Rightarrow \left( {x_1^2 + x_2^2 + x_3^2} \right) \ge \left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)\\ \Rightarrow \left( {x_1^2 + x_2^2 + x_3^2} \right) + 2\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right) \ge 3\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)\\ \Rightarrow {\left( {{x_1} + {x_2} + {x_3}} \right)^2} \ge 3\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)\end{array}\)

\(\left( * \right) \Leftrightarrow x_1^7 + x_2^7 + x_3^7 \ge \frac{1}{{729}}{\left( {{x_1} + {x_2} + {x_3}} \right)^7}\)(**)

Áp dụng BĐT Bunhiacopxki ta có:

\(\begin{array}{l}\,\,\,\,\,\left( {x_1^7 + x_2^7 + x_3^7} \right)\left( {{x_1} + {x_2} + {x_3}} \right) \ge \frac{1}{{81}}{\left( {{x_1} + {x_2} + {x_3}} \right)^3}{\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)^2}\\ \Rightarrow \left( {x_1^7 + x_2^7 + x_3^7} \right)\left( {{x_1} + {x_2} + {x_3}} \right) \ge {\left( {x_1^4 + x_2^4 + x_3^4} \right)^2}\\ \Rightarrow \left( {x_1^4 + x_2^4 + x_3^4} \right)\left( {1 + 1 + 1} \right) \ge {\left( {x_1^2 + x_2^2 + x_3^2} \right)^2}\\ \Rightarrow {\left( {x_1^4 + x_2^4 + x_3^4} \right)^2} \ge \frac{{{{\left( {x_1^2 + x_2^2 + x_3^2} \right)}^4}}}{9}\\\,\,\,\,\,\,\left( {x_1^2 + x_2^2 + x_3^2} \right) \ge \frac{{{{\left( {{x_1} + {x_2} + {x_3}} \right)}^2}}}{3}\\ \Rightarrow {\left( {x_1^2 + x_2^2 + x_3^2} \right)^4} \ge \frac{{{{\left( {{x_1} + {x_2} + {x_3}} \right)}^4}}}{{81.9}}\\ \Rightarrow \left( {x_1^7 + x_2^7 + x_3^7} \right)\left( {{x_1} + {x_2} + {x_3}} \right) \ge \frac{{{{\left( {{x_1} + {x_2} + {x_3}} \right)}^4}}}{{729}}\\ \Rightarrow \left( {x_1^7 + x_2^7 + x_3^7} \right) \ge \frac{{{{\left( {{x_1} + {x_2} + {x_3}} \right)}^3}}}{{729}}\end{array}\)

Suy ra (**) đúng

Suy ra \( \Leftrightarrow x_1^7 + x_2^7 + x_3^7 \ge \frac{1}{{81}}{\left( {{x_1} + {x_2} + {x_3}} \right)^3}{\left( {{x_1}{x_2} + {x_1}{x_3} + {x_3}{x_2}} \right)^2}\)

Hay \(x_1^7 + x_2^7 + x_3^7 \ge \frac{{ - {b^3}.{c^2}}}{{81{a^5}}}\) (đpcm).