Topic Inequality THPT
#1
Posted 05-05-2011 - 17:59
1. Cho a,b và c là các số thực dương sao cho abc=1. CMR:
$\dfrac{{ab}}{{a^5 + b^5 + ab}} + \dfrac{{bc}}{{b^5 + c^5 + bc}} + \dfrac{{ca}}{{c^5 + a^5 + ca}} \le 1$
2. Cho các số thực dương a,b,c. Cmr:
$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge \dfrac{{c + a}}{{c + b}} + \dfrac{{a + b}}{{a + c}} + \dfrac{{b + c}}{{b + a}}$
3. Cho a,b,c,d >0 sao cho $a \le b \le c \le d$ và abcd=1. CMR:
$\left( {a + 1} \right)\left( {d + 1} \right) \ge 3 + \dfrac{3}{{4d^3 }}$
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
Posted 05-05-2011 - 19:26
Đây sẽ là chủ đề thảo luận, hỏi về các BĐT của THPT, mở đầu xin giới thiệu một số bài toán
1. Cho a,b và c là các số thực dương sao cho abc=1. CMR:
$\dfrac{{ab}}{{a^5 + b^5 + ab}} + \dfrac{{bc}}{{b^5 + c^5 + bc}} + \dfrac{{ca}}{{c^5 + a^5 + ca}} \le 1$
2. Cho các số thực dương a,b,c. Cmr:
$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge \dfrac{{c + a}}{{c + b}} + \dfrac{{a + b}}{{a + c}} + \dfrac{{b + c}}{{b + a}}$
3. Cho a,b,c,d >0 sao cho $a \le b \le c \le d$ và abcd=1. CMR:
$\left( {a + 1} \right)\left( {d + 1} \right) \ge 3 + \dfrac{3}{{4d^3 }}$
mình chém bài 1:
Áp dụng $\ x^5+y^5\ge\xy(x^3+y^3)$
VT$\le\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}$
Mà:$\ x^3+y^3\ge\xy(x+y)$
VT$\le\dfrac{1}{ab(a+b)+1}+\dfrac{1}{bc(b+c)+1}+\dfrac{1}{ca(c+a)+1}$=
$\dfrac{1}{ab(a+b)+abc}+\dfrac{1}{bc(b+c)+abc}+\dfrac{1}{ca(c+a)+abc}$
Đến đây thì OK rùi nhik
Bài 2: Trừ mỗi vế cho ba rồi chuyển vế biến đổi tương đương chắc ra đó!!
Edited by harrypotter10a1, 05-05-2011 - 19:33.
#3
Posted 05-05-2011 - 19:42
hình như pạn nhầm chỗmình chém bài 1:
Áp dụng $\ x^5+y^5\ge\xy(x^3+y^3)$
VT$\le\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}$
Mà:$\ x^3+y^3\ge\xy(x+y)$
VT$\le\dfrac{1}{ab(a+b)+1}+\dfrac{1}{bc(b+c)+1}+\dfrac{1}{ca(c+a)+1}$=
$\dfrac{1}{ab(a+b)+abc}+\dfrac{1}{bc(b+c)+abc}+\dfrac{1}{ca(c+a)+abc}$
Đến đây thì OK rùi nhik
Bài 2: Trừ mỗi vế cho ba rồi chuyển vế biến đổi tương đương chắc ra đó!!
$\begin{array}{l}{x^5} + {y^5} \ge {x^3} + {y^3}\left( {sai} \right)\\{x^5} + {y^5} \ge {x^2}{y^2}\left( {x + y} \right)\end{array}$
mới đúng
Edited by Lê Xuân Trường Giang, 05-05-2011 - 19:43.
N.HÍCHMÉT
Khó + Lười = Bất lực
#4
Posted 06-05-2011 - 09:59
$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge \dfrac{{c + a}}{{c + b}} + \dfrac{{a + b}}{{a + c}} + \dfrac{{b + c}}{{b + a}}$
Ta có :
$\sum \dfrac{a}{b} -3= \dfrac{(a-b)^2}{ab} + \dfrac{(a-c)(b-c)}{ac} $
BDT được viết lại thành :
$\ [ \dfrac{1}{ab} - \dfrac{1}{(a+c)(b+c)} ](a-b)^2+[ \dfrac{1}{ac} - \dfrac{1}{(a+c)(b+a)} ](a-c)(b-c) \geq 0 $
Chỉ cần giả sử $\ c=min {a,b,c} $ là ta có ngay đpcm
Đã xong . . .
P/s : Bài toán tổng quát như sau :
Cho $\ a,b,c,k $ là các số thực dương . CMR:
$\sum \dfrac{a}{b} \geq \sum \dfrac{a+kb}{a+kc} $
Xăng có thể cạn, lốp có thể mòn..xong số máy số khung thì không bao giờ thay đổi
NGUYỄN ANH TUẤN - CHỦ TỊCH HIỆP HỘI
#5
Posted 06-05-2011 - 14:29
1. Cho a,b,c là các cố thực dương, CMR:
$\dfrac{a}{{\left( {b + c} \right)^2 }} + \dfrac{b}{{\left( {c + a} \right)^2 }} + \dfrac{c}{{\left( {a + b} \right)^2 }} \ge \dfrac{9}{{4\left( {a + b + c} \right)}}$
2. Cho a,b,c > 0 và a+b+c = 1. CMR: $7\left( {ab + bc + ca} \right) \le 2 + 9abc$
3. Cho $a,b,c \in R _ + $. CMR:
$\sqrt {a\left( {b + 1} \right)} + \sqrt {b\left( {c + 1} \right)} + \sqrt {c\left( {a + 1} \right)} \le \dfrac{3}{2}\sqrt {\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)} $
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”).
#6
Posted 06-05-2011 - 15:31
Không mqaats tính tổng quát, giả sử $a \ge b \ge c \to a \le \dfrac{1}{3}$
ta cần chứng minh: $bc(7-9a) + 7a(1-a) \le 2$
hiển nhiên: $bc \le \dfrac{(b+c)^2}{4} = \dfrac{(1-a)^2}{4}$ và $7-9a > 0$
nên $VT \le (7-9a)\dfrac{(1-a)^2}{4} + 7a(1-a) \le 2$
đến đây có thể quy về xét hàm số $f(a)$ với $a \in (0; \dfrac{1}{3}]$ hoặc phân tichs tương đương thành nhân tử cũng ok!
p/s: có thể dùng các cách khác như BDT Schur, xét khoảng đánh giá ,.....
Edited by h.vuong_pdl, 06-05-2011 - 15:33.
rongden_167
#7
Posted 06-05-2011 - 18:44
$\dfrac{a}{{\left( {b + c} \right)^2 }} + \dfrac{b}{{\left( {c + a} \right)^2 }} + \dfrac{c}{{\left( {a + b} \right)^2 }} \ge \dfrac{9}{{4\left( {a + b + c} \right)}}$
chém bài này nhé
đặt a+b=x, b+c=y, c+a=z (x,y,z>0)
$ BDT \Leftrightarrow \dfrac{(x+z)^2-y^2}{y^2}+\dfrac{(x+y)^2-z^2}{z^2} + \dfrac{(y+z)^2-x^2}{x^2} \geq 9 $.
$ \Leftrightarrow \dfrac{(z+x)^2}{y^2}+\dfrac{(x+y)^2}{z^2}+\dfrac{(y+z)^2}{x^2} \geq 12 $
đây là điều hiển nhiên đúng theo BDT AM-GM
xong!
Edited by NGOCTIEN_A1_DQH, 06-05-2011 - 19:55.
Mong rằng toán học bớt khô khan
Em ơi trong toán nhiều công thức
Cũng đẹp như hoa lại chẳng tàn
#8
Posted 06-05-2011 - 19:02
CMR
$\dfrac{\sqrt{ab+4bc+4ca}}{a+b} +\dfrac{\sqrt{bc+4ac+ab}}{b+c} +\dfrac{\sqrt{ca+4ab+4bc}}{c+a} \geq \dfrac{9}{2}$
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
#9
Posted 06-05-2011 - 20:48
$\\\sqrt {a(b + 1)} + \sqrt {b(c + 1)} + \sqrt {c(a + 1)} \le \dfrac{3}{2}\sqrt {(a + 1)(b + 1)(c + 1)} \\\\\Leftrightarrow \sqrt {\dfrac{a}{{a + 1}}.\dfrac{1}{{c + 1}}} + \sqrt {\dfrac{b}{{b + 1}}.\dfrac{1}{{a + 1}}} + \sqrt {\dfrac{c}{{c + 1}}.\dfrac{1}{{b + 1}}} \le \dfrac{3}{2}\\\\VT \le \dfrac{1}{2}\left[ {\dfrac{a}{{a + 1}} + \dfrac{1}{{c + 1}}} \right] + \dfrac{1}{2}\left[ {\dfrac{b}{{b + 1}} + \dfrac{1}{{a + 1}}} \right] + \dfrac{1}{2}\left[ {\dfrac{c}{{c + 1}} + \dfrac{1}{{b + 1}}} \right] = \dfrac{3}{2} \\$
dấu = xảy ra khi và chỉ khi a=b=c=1
Edited by truclamyentu, 06-05-2011 - 20:52.
#10
Posted 06-05-2011 - 21:05
cm bất đẳng thức sau với mọi a,b,c>0
$\left[ {a + 1- \dfrac{1}{b}} \right]\left[ {b + 1 -\dfrac{1}{c}}\right] +\left[ {b + 1 -\dfrac{1}{c}}\right]\left[{c + 1 -\dfrac{1}{a}} \right] + \left[ {c + 1 - \dfrac{1}{a}} \right]\left[ {a + 1 - \dfrac{1}{b}} \right] \ge3$
Edited by truclamyentu, 06-05-2011 - 21:07.
#11
Posted 08-05-2011 - 09:55
Xin hỏi a,b,c là các số như thế nào bạn?góp vui một bài!
CMR
$\dfrac{\sqrt{ab+4bc+4ca}}{a+b} +\dfrac{\sqrt{bc+4ac+ab}}{b+c} +\dfrac{\sqrt{ca+4ab+4bc}}{c+a} \geq \dfrac{9}{2}$
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”).
#12
Posted 18-05-2011 - 20:00
$\left ( a + b \right )\left ( c + d \right )+\left ( a + d \right )\left ( b+c \right )\geq 4\sqrt{\left ( 1 + ac \right )\left ( 1 + bd \right )}$
Problem 2: Cho $a\geq b\geq c> 0$.CMR:
$\dfrac{a^{2}-b^{2}}{c}+\dfrac{c^{2}-b^{2}}{a}+\dfrac{a^{2}-c^{2}}{b}\geq 3a-4b+c$
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”).
#13
Posted 18-05-2011 - 21:20
Problem 1: Cho các số thực dương $a,b,c,d$ sao cho $a + b + c + d = abc + abd + acd + bcd$.CMR:
$\left ( a + b \right )\left ( c + d \right )+\left ( a + d \right )\left ( b+c \right )\geq 4\sqrt{\left ( 1 + ac \right )\left ( 1 + bd \right )}$
Problem 2: Cho $a\geq b\geq c> 0$.CMR:
$\dfrac{a^{2}-b^{2}}{c}+\dfrac{c^{2}-b^{2}}{a}+\dfrac{a^{2}-c^{2}}{b}\geq 3a-4b+c$
PROBLEM2/
$(a + c) + (b + d) = (abc + acd) + (abd + bcd)\\\\ \Leftrightarrow (a + c) + (b + d) = ac(b + d) + bd(a + c) \le \dfrac{1}{4}{(a + c)^2}(b + d) + \dfrac{1}{4}{(b + d)^2}(a + c)\\\\ \Rightarrow (a + c) + (b + d) \le \dfrac{1}{4}{(a + c)^2}(b + d) + \dfrac{1}{4}{(b + d)^2}(a + c)\\\\ \Leftrightarrow (a + c)(b + d) \ge 4\\\\ \Leftrightarrow (a + b)(c + d) + (a + d)(b + c) \ge 2(ac + 1 + bd + 1)\\\\2(ac + 1 + bd + 1) \ge 4\sqrt {(ac + 1)(bd + 1)} \\\\\Rightarrow (a + b)(c + d) + (a + d)(b + c) \ge 4\sqrt {(ac + 1)(bd + 1)} $
Edited by truclamyentu, 18-05-2011 - 21:23.
#14
Posted 20-05-2011 - 18:22
Bài toán này là từ cuốn của Xuezhi Yang, ta có thể sử dụng BĐT Iran 96 để giải bài này! Bạn có thể xem rõ hơn ở đây:góp vui một bài!
CMR
$\dfrac{\sqrt{ab+4bc+4ca}}{a+b} +\dfrac{\sqrt{bc+4ac+ab}}{b+c} +\dfrac{\sqrt{ca+4ab+4bc}}{c+a} \geq \dfrac{9}{2}$
http://www.artofprob...v...51&t=407329
Problem 3: Let $a, b, c$ denote the lengths of the sides of a triangle, and let $u, v, w$, respectively, be the distances of the centre of the incircle from the vertices opposite to the sides. Prove that
$\left ( a + b + c \right )\left ( \dfrac{1}{u} + \dfrac{1}{v} + \dfrac{1}{w} \right )\leq 3\left ( \dfrac{a}{u} + \dfrac{b}{v} + \dfrac{c}{w} \right )$
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”).
#15
Posted 20-05-2011 - 18:32
Link:http://diendantoanho...mp;#entry261478
#16
Posted 20-05-2011 - 20:03
Cảm ơn anh!
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”).
#17
Posted 20-05-2011 - 20:09
Không cần thiết phải như vậy vì trên diễn đàn chỉ cần một topic về chuyên đề BĐT thôi,làm như vậy sẽ tránh làm loãng topic của cả 2 bên.Thân.Thiệt tình em cũng thấy anh hơi vô lí. Topic của anh wallunint là topic BĐT của phần Olympiad, còn em tổ chức ở phần THPT cơ mà! Mỗi topic đều có điểm khác nhau của nó chứ! Đâu phải topic của em hoàn toàn giống của anh wallunint. Em hi vọng topic này sẽ được tiếp tục!
Cảm ơn anh!
P/s:Anh vẫn sẽ đóng topic này lại.
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