Chứng minh rằng nếu $a,b,c>0$ thì \[\frac{1}{2} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ac}} \ge \frac{a}{{b + c}} + \frac{b}{{a + c}} + \frac{c}{{a + b}} \ge (4 - \frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}})\]
CM $\frac{a}{{b + c}} + \frac{b}{{a + c}} + \frac{c}{{a + b}} \ge (4 - \frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}})$
Bắt đầu bởi Ispectorgadget, 26-06-2012 - 23:07
#1
Đã gửi 26-06-2012 - 23:07
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#2
Đã gửi 27-06-2012 - 09:40
Mấy bài này cũ rồi dành cho THCS thôi à!Chứng minh rằng nếu $a,b,c>0$ thì \[\frac{1}{2} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ac}} \ge \frac{a}{{b + c}} + \frac{b}{{a + c}} + \frac{c}{{a + b}} \ge (4 - \frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}})\]
Đề như thế này mới đúng nè:
\[\frac{1}{2} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ac}} \ge \frac{a}{{b + c}} + \frac{b}{{a + c}} + \frac{c}{{a + b}} \ge (2 - \frac{1}{2}\frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}})\]
Đặt $\frac{a}{{b + c}} + \frac{b}{{a + c}} + \frac{c}{{a + b}}$
Ta có:
$P \ge \frac{(a+b+c)^2}{2(ab+bc+ac)}=1+\frac{{{a^2} + {b^2} + {c^2}}}{{2(ab + bc + ac)}}$
Do đó:
$P+\frac{1}{2}\frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}} \ge 1+ \frac{1}{2}\frac{{ab + bc + ac}}{{{a^2} + {b^2} + {c^2}}}+\frac{{{a^2} + {b^2} + {c^2}}}{{2(ab + bc + ac)}}$
$\\ge 2(AM-GM)$
Còn vế sau:
$P \le \frac{1}{2} + \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ac}}$
$\Leftrightarrow (\sum ab)(\sum \frac{a}{b+c}) \le \frac{\sum ab}{2}+\sum a^2$
$\Leftrightarrow abc(\sum \frac{1}{b+c}) \le \frac{\sum ab}{2}$
Mà $\sum \frac{4}{b+c} \le 2(\sum \frac{1}{a})$
Hay : $2(\sum \frac{1}{b+c}) \le \sum \frac{1}{a}=\frac{\sum ab}{abc}$
Suy ra : $Q.E.D$
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