$\dfrac{ 1}{2a^2+bc} + \dfrac{1}{2b^2+ca} + \dfrac{1}{2c^2+ab} \leq \dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)^2$
Bài viết đã được chỉnh sửa nội dung bởi dark templar: 22-05-2011 - 20:04
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#1
Đã gửi 16-04-2011 - 06:09
Zaraki
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PQT
Cho $a,b,c$ là các số thực dương. CMR
$\dfrac{ 1}{2a^2+bc} + \dfrac{1}{2b^2+ca} + \dfrac{1}{2c^2+ab} \leq \dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)^2$
$\dfrac{ 1}{2a^2+bc} + \dfrac{1}{2b^2+ca} + \dfrac{1}{2c^2+ab} \leq \dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)^2$
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#2
Đã gửi 16-04-2011 - 10:34
phuc_90
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Cho a,b,c là các số thực dương. CMR
$ \dfrac{1}{2a^2+bc} + \dfrac{1}{2b^2+ca} + \dfrac{1}{2c^2+ab} \leq \dfrac{1}{9} (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2 $
Sử dụng bđt Cauchy 3 số ở mẫu là ra rồi bạn !!
Bài viết đã được chỉnh sửa nội dung bởi phuc_90: 16-04-2011 - 10:43
#3
Đã gửi 16-04-2011 - 12:59
Zaraki
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Anh ghi luôn lời giải ra được không ?
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#4
Đã gửi 16-04-2011 - 14:47
tuan101293
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Trung úy
ta cóSử dụng bđt Cauchy 3 số ở mẫu là ra rồi bạn !!
$\dfrac{9}{2a^2+bc}\le \dfrac{9}{3\sqrt[3]{a^4bc}}=\dfrac{3}{\sqrt[3]{a^4bc}}\le \dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{ac}$
Làm tương tự 2 cái cộng vào là ra
KT-PT
Do unto others as you would have them do unto you.
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