Chủ đề này có 5 trả lời

[sonksnb]

Cho a,b,c là những số dương.Tìm GTNN:
$\frac{(a+b+c)^{3}}{abc}+(\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}})^{2}$

[truclamyentu]

Cho a,b,c là những số dương.Tìm GTNN:
$\frac{(a+b+c)^{3}}{abc}+(\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}})^{2}$

Cách 1:dùng SOS
Ta có :
$A = \frac{{{{(a + b + c)}^3}}}{{abc}} - 27 + {\left( {\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}}} \right)^2} - 1$

$ = \frac{{\left( {\sum\limits_{cyc} {{a^3}} - 3abc} \right) + \left( {3\sum\limits_{sym} {{a^2}b} - 18abc} \right)}}{{abc}} - \frac{{\sum\limits_{cyc} {{{(b - c)}^2}} \left( {\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} } \right)}}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}$

$ = \frac{{\frac{1}{2}\sum\limits_{cyc} a \sum\limits_{cyc} {{{(b - c)}^2}} + 3\sum\limits_{cyc} {\left( {a{{(b - c)}^2}} \right)} }}{{abc}} - \frac{{\sum\limits_{cyc} {{{(b - c)}^2}} \left( {\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} } \right)}}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}$

$ = \sum\limits_{cyc} ( b - c{)^2}\left[ {\frac{{\frac{1}{2}\sum\limits_{cyc} a + 3a}}{{abc}} - \frac{{\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} }}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}} \right]$
Ta đi chứng minh :

$\frac{{\frac{1}{2}\sum\limits_{cyc} a + 3a}}{{abc}} - \frac{{\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} }}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}\ge0 $

$ \Leftrightarrow \frac{{\frac{1}{2}\sum\limits_{cyc} a + 3a}}{{abc}} \ge \frac{{\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} }}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}$
Chuẩn hóa :$\sum\limits_{cyc} {bc} = 3 \Rightarrow \sum\limits_{cyc} {{a^2}} \ge 3;\sum\limits_{cyc} a \ge 3;abc \le 1$
Khi đó :
$\frac{{\frac{1}{2}\sum\limits_{cyc} a + 3a}}{{abc}} > \frac{{\frac{1}{2}\sum\limits_{cyc} a }}{{abc}} \ge 1,5 > \frac{1}{3} > \frac{1}{{2.3}} + \frac{3}{{2.9}} \ge \frac{1}{{2\sum\limits_{cyc} {{a^2}} }} + \frac{{\sum\limits_{cyc} {bc} }}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}} = \frac{{\sum\limits_{cyc} {{a^2}} + \sum\limits_{cyc} {bc} }}{{2{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}$
Vấy min A =28 khi a=b=c

Bài viết đã được chỉnh sửa nội dung bởi truclamyentu: 14-05-2012 - 19:49

[truclamyentu]

Cách 2 : Dùng cauchy

Ta có :
$\frac{{{{(a + b + c)}^3}}}{{abc}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2} \ge \frac{{9{{(a + b + c)}^3}}}{{(a + b + c)(ab + bc + ca)}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$= \frac{{9{{(a + b + c)}^2}}}{{ab + bc + ca}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$ = \frac{{9({a^2} + {b^2} + {c^2}) + 18(ab + bc + ca)}}{{ab + bc + ca}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$ = \left( {\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + {{(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})}^2}} \right) + 7.\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + 18$

$ \ge 3 + 7 + 18 = 28$

Bài viết đã được chỉnh sửa nội dung bởi truclamyentu: 14-05-2012 - 20:04

[sonksnb]

Cách 2 : Dùng cauchy

Ta có :
$\frac{{{{(a + b + c)}^3}}}{{abc}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2} \ge \frac{{9{{(a + b + c)}^3}}}{{(a + b + c)(ab + bc + ca)}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$= \frac{{9{{(a + b + c)}^2}}}{{ab + bc + ca}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$ = \frac{{9({a^2} + {b^2} + {c^2}) + 18(ab + bc + ca)}}{{ab + bc + ca}} + {(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})^2}$

$ = \left( {\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + {{(\frac{{ab + bc + ca}}{{{a^2} + {b^2} + {c^2}}})}^2}} \right) + 7.\frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} + 18$

$ \ge 3 + 7 + 18 = 28$

Bạn ơi sao$\frac{(a+b+c)^{3}}{abc}\geq \frac{9(a+b+c)^{3}}{(a+b+c)(ab+bc+ca)}$

[truclamyentu]

Bạn ơi sao$\frac{(a+b+c)^{3}}{abc}\geq \frac{9(a+b+c)^{3}}{(a+b+c)(ab+bc+ca)}$

Bởi vì $(a+b+c)(ab+bc+ca)\ge 9abc$

[sonksnb]

Bởi vì $(a+b+c)(ab+bc+ca)\ge 9abc$

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