Bài toán trong sách tài liệu chuyên Toán
#1
Đã gửi 13-09-2011 - 20:03
$\sum\limits_{i = 1}^n {\dfrac{1}{{1 + x_i^2}}} = 1$
Chứng minh rằng
$\sum\limits_{i = 1}^n {{x_i}} \ge \left( {n - 1} \right)\sum\limits_{i = 1}^n {\dfrac{1}{{{x_i}}}} $
#2
Đã gửi 20-09-2011 - 20:55
Bất đẳng thức trở thành :
$\sum_{i=1}^{n}\sqrt{\dfrac{1-a_i}{a_i}}\geq(n-1)(\sum_{i=1}^{n}\sqrt{\dfrac{a_i}{1-a_i}})$
Thật vậy :
$LHS= \sum_{i=1}^{n}\sqrt{\dfrac{a_1+a_2+...+a_{i-1}+a_{i+1}+..+a_n}{a_i}}$
$\overset {Cauchy-Schwarz}{\ge}\sum_{i=1}^{n}\dfrac{\sqrt{a_1}+...+\sqrt{a_{i-1}}+\sqrt{a_{i+1}}+...\sqrt{a_n}}{\sqrt{n-1}\sqrt{a_i}}$
$=\sum_{i=1}^{n}\dfrac{\sqrt{a_i}}{\sqrt{n-1}}.(\dfrac{1}{\sqrt{a_1}}+\dfrac{1}{\sqrt{a_2}}+...+\dfrac{1}{\sqrt{a_{i-1}}}+\dfrac{1}{\sqrt{a_{i+1}}}+...+\dfrac{1}{\sqrt{a_n}})$
$\geq \dfrac{(n-1)\sqrt{a_i}}{\sqrt{\sum_{i=1}^{n}a_i}-a_i}=(n-1)(\sum_{i=1}^{n}\sqrt{\dfrac{a_i}{1-a_i}})$
$Q.E.D$
Bài viết đã được chỉnh sửa nội dung bởi PRONOOBCHICKENHANDSOME: 20-09-2011 - 21:05
- NguyThang khtn và Zaraki thích
#3
Đã gửi 30-09-2011 - 16:42
#4
Đã gửi 30-09-2011 - 17:45
Đây là viết tắt của tiếng anh (đôi lúc dùng cho hay hay).LHS va Q.E.D la cai gi vay ban? minh doc ko hieu cai do la cai gi.
LHS tức là vế trái.
Q.E.D tức là đpcm.
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”).
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