Chủ đề này có 1 trả lời

[xuantungjinkaido]

$a,b,c>0$ thỏa mãn $a+b+c=1$ tìm min $Q=ab+bc+ca+\frac{3}{2}\left ( \frac{1}{a} +\frac{1}{b}+\frac{1}{c}\right )$

Bài viết đã được chỉnh sửa nội dung bởi tpdtthltvp: 15-05-2016 - 22:52

[PlanBbyFESN]

$a,b,c>0$ thỏa mãn $a+b+c=1$ tìm min $Q=ab+bc+ca+\frac{3}{2}\left ( \frac{1}{a} +\frac{1}{b}+\frac{1}{c}\right )$

 

AM-GM: $(ab+bc+ca)^{2}\geq 3abc(a+b+c)$

 

AM-GM & Cauchy-Schwarz:

 

$ab+bc+ca+\frac{3}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=ab+bc+ca+\frac{1}{27}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{79}{54}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 2\sqrt{\frac{1}{27}(ab+bc+ca)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}+\frac{79}{54}(\frac{9}{a+b+c})=2\sqrt{\frac{1}{27}\frac{(ab+bc+ca)^{2}}{abc}}+\frac{79}{6}\geq 2\sqrt{\frac{1}{27}\frac{3abc(a+b+c)}{abc}}+\frac{79}{6}=\frac{83}{6}$

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