$\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge\sqrt{82}$
$\sum \sqrt{x^{2}+\dfrac{1}{x^{2}}}\geq\sqrt{82}$
#1
Đã gửi 08-08-2011 - 19:25
- hxthanh và minhdat881439 thích
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 08-08-2011 - 19:28
Bài này là đề thi đại học khối A năm 2003 mà.Cho $x,y,z \ge 0, x+y+z \le 1$. Chứng minh rằng
$\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge\sqrt{82}$
Mình xin giới thiệu một cách;
Ta có: $\left( {x + y + z} \right)^2 + \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)^2 = 81\left( {x + y + z} \right)^2 + \left({\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)^2 - 80\left( {x + y + z} \right)^2 $
$\ge 18\left( {x + y + z} \right) + \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) - 80\left( {x + y + z} \right)^2 \ge 162 - 80 = 82$
Vậy $VT \ge \sqrt {82} $
Đẳng thức xảy ra khi $x = y = z = \dfrac{1}{3}$.
P/S: Bài này có nhiều cách giải khác.
Bài viết đã được chỉnh sửa nội dung bởi xusinst: 08-08-2011 - 19:35
#3
Đã gửi 08-08-2011 - 19:39
Áp dụng BĐT Minkowski, ta có
$\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge\sqrt{(x+y+z)^{2}+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^{2}}\ge\sqrt{(x+y+z)^{2}+\dfrac{81}{(x+y+z)^{2}}}=\sqrt{(x+y+z)^{2}+\dfrac{1}{(x+y+z)^{2}}+\dfrac{80}{(x+y+z)^{2}}}\ge\sqrt{2+\dfrac{80}{(x+y+z)^{2}}}\ge\sqrt{82}$
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 08-08-2011 - 19:46
$\sum \sqrt{x^{2}+ \dfrac{1}{x^{2} } \ge \sqrt{(x+y+z)^{2}+( \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z})^{2} } \ge \sqrt{(x+y+z)^{2}+ \dfrac{81}{(x+y+z)^{2}} } \ge \sqrt{(x+y+z)^{2}+ \dfrac{1}{(x+y+z)^{2}} + \dfrac{80}{(x+y+z)^{2}}} \ge \sqrt{82}$ (bài toàn đc giải bằng minouski và cauchy)Cho $x,y,z \ge 0, x+y+z \le 1$. Chứng minh rằng
$\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge\sqrt{82}$
#5
Đã gửi 08-08-2011 - 19:47
$ A=\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge 3\sqrt[6]{(x^{2}+\dfrac{1}{x^{2}})(y^{2}+\dfrac{1}{y^{2}})(z^{2}+\dfrac{1}{z^{2}})} $.
Ta có
$x^{2}+\dfrac{1}{x^{2}}=x^{2}+\underset{81}{\underbrace{\dfrac{1}{81x^{2}}+\dfrac{1}{81x^{2}}+...+\dfrac{1}{81x^{2}}}}\geq 82\sqrt[82]{\dfrac{1}{81^{81}x^{160}}} $.
Tương tự
$ y^{2}+\dfrac{1}{y^{2}}\geq 82\sqrt[82]{\dfrac{1}{81^{81}y^{160}}} $.
$ z^{2}+\dfrac{1}{z^{2}}\geq 82\sqrt[82]{\dfrac{1}{81^{81}z^{160}}} $.
$ \Rightarrow A=\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}} $
$ \ge 3\sqrt[6]{(x^{2}+\dfrac{1}{x^{2}})(y^{2}+\dfrac{1}{y^{2}})(z^{2}+\dfrac{1}{z^{2}})} $
$ \geq 3\sqrt[6]{82^{3}\sqrt[82]{\dfrac{1}{81^{243}(xyz)^{160}}}} $
Theo giả thiết thì $x+y+z\leq 1\Rightarrow xyz\leq\dfrac{1}{27}$
$ \Rightarrow 3\sqrt[6]{82^{3}\sqrt[82]{\dfrac{1}{81^{243}(xyz)^{160}}}}\geq 3\sqrt[6]{82^{3}\sqrt[82]{\dfrac{1}{81^{243}(\dfrac{1}{27})^{160}}}}=\sqrt{82} $.
$ \boxed{A\ge 3\sqrt[6]{(x^{2}+\dfrac{1}{x^{2}})(y^{2}+\dfrac{1}{y^{2}})(z^{2}+\dfrac{1}{z^{2}})}\geq 3\sqrt[6]{82^{3}\sqrt[82]{\dfrac{1}{81^{243}(xyz)^{160}}}}\geq\sqrt{82}} $.
Đẳng thức xảy ra khi $x=y=z= \dfrac{1}{3}$.
- Tru09 yêu thích
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
Đã gửi 08-08-2011 - 19:47
Sao cái căn này nó dài thế$\sum \sqrt{x^{2}+ \dfrac{1}{x^{2} } \ge \sqrt{(x+y+z)^{2}+( \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z})^{2} } \ge \sqrt{(x+y+z)^{2}+ \dfrac{81}{(x+y+z)^{2}} } \ge \sqrt{(x+y+z)^{2}+ \dfrac{1}{(x+y+z)^{2}} + \dfrac{80}{(x+y+z)^{2}}} \ge \sqrt{82}$ (bài toàn đc giải bằng minouski và cauchy)
#7
Đã gửi 08-08-2011 - 19:49
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”).
#8
Đã gửi 13-04-2012 - 22:45
Thêm một cách nữaCho $x,y,z \ge 0, x+y+z \le 1$. Chứng minh rằng
$\sqrt{x^{2}+\dfrac{1}{x^{2}}}+\sqrt{y^{2}+\dfrac{1}{y^{2}}}+\sqrt{z^{2}+\dfrac{1}{z^{2}}}\ge\sqrt{82}$
Xét 3 vecto: $\overrightarrow{a}(x;\frac{1}{y}), \overrightarrow{b}(y;\frac{1}{z}), \overrightarrow{c}(z;\frac{1}{x})$
$\Rightarrow (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) (x+y+z;\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
Áp dụng công thức :
$\left | \overrightarrow{a} \right |+\left | \overrightarrow{b} \right |+\left | \overrightarrow{c} \right |\geq \left | \overrightarrow{a} +\overrightarrow{b}+\overrightarrow{c}\right |$
ta có :
$VT\geq \sqrt{(x+y+z)^{2}+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}}\geq \sqrt{(x+y+z)^{2}+(\frac{9}{x+y+z})^{2}}= \sqrt{82}$
#9
Đã gửi 26-07-2012 - 21:18
Cho $x, y, z > 0$ và $x +y +z \leq 2$ . CMR:
$\sqrt{x^{2}+ \frac{1}{y_{2}}}$ + $\sqrt{y^{2}+ \frac{1}{z_{2}}}$ + $\sqrt{z^{2}+ \frac{1}{x_{2}}}$ $\geq$ $\sqrt{\frac{97}{2}}$
#10
Đã gửi 26-07-2012 - 21:27
Nãy em đang post thì không ngờ topic bị khóa.Thôi qua đây vậyBài toán của bạn jackboy225
Áp dụng BĐT Bunyakovsky,ta có:
$\sqrt{(x^2+\frac{1}{y^2})97}=\sqrt{(9^2+4^2)(x^2+\frac{1}{y^2})}\geq 4x+\frac{9}{y}$
Thiết lập các BĐT tương tự rồi cộng vế theo vế,ta được:
$\sum \sqrt{97}\sqrt{x^2+\frac{1}{y^2}}\geq 4x+4y+4z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}$
$=(\frac{81}{4}x+\frac{9}{x})+(\frac{81}{4}y+\frac{9}{y})+(\frac{81}{4}z+\frac{9}{z})-\frac{65}{4}(x+y+z)\geq$
$ 27+27+27-\frac{65}{2}(cauchy,x+y+z\leq 2)=\frac{97}{2}$
$\Rightarrow \sum \sqrt{x^2+\frac{1}{y^2}}\geq \frac{\sqrt{97}}{2}(Q.E.D)$
P/s:Theo như mình xem qua thì chưa ai giải bài này bằng cách dùng Bunyakovsky,mình xin giới thiệu cách mới.Đây là lần đầu tiên mình giải 1 bài toán trong box Olympic,mọi người ủng hộ nhé!!!
Bài viết đã được chỉnh sửa nội dung bởi triethuynhmath: 26-07-2012 - 21:29
- hxthanh và minhdat881439 thích
TRIETHUYNHMATH
___________________________
08/12/1997
#11
Đã gửi 26-07-2012 - 21:35
Bước 1. Áp dụng BĐT Minkopski như bài giải số 2.
Bước 2. Khảo sát hàm $f(t)=t^2+\frac{81}{t^2}$ trên $(- \infty ; 2]$
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