Cho a,b,c>0 va a+b+c=0.Tim GTLN:
A=(a^3+b^3+c^3)/abc
BDT de
Bắt đầu bởi tunghieu, 30-10-2005 - 10:24
#1
Đã gửi 30-10-2005 - 10:24
#2
Đã gửi 30-10-2005 - 10:35
Thật sự mình không hiểu bạn định viết gì.Cho a,b,c>0 va a+b+c=0.
Trying not to break
#3
Đã gửi 30-10-2005 - 12:59
sao a, b, c>0 mà a+b+c=0 được
Kiên nhẫn là nghệ thuật của hi vọng. Tài năng là 1% còn 99% là mồ hôi nước mắt
#4
Đã gửi 30-10-2005 - 17:54
Chắc bạn đó type nhầm
Tôi nghĩ đề là :
Tìm max :
http://dientuvietnam.net/cgi-bin/mimetex.cgi?P=\dfrac{x^3+y^3+z^3}{xyz}
biết http://dientuvietnam.net/cgi-bin/mimetex.cgi?P=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}
G/s http://dientuvietnam...imetex.cgi?(x-y)(x^2-y^2)\geq{0}
<=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?xy^2+yz^2-xz^2\geq{y^3}
<= http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{y}{z}+\dfrac{z}{x}-\dfrac{z}{y}\geq{\dfrac{x^2}{yz}} (1)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{x^2}{yz}\leq{\dfrac{x}{z}} (do http://dientuvietnam.net/cgi-bin/mimetex.cgi?1\leq{x}\leq{y} ) (2)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{z^2}{xy}\leq{\dfrac{2z}{y}} (do http://dientuvietnam.net/cgi-bin/mimetex.cgi?1\leq{x} và http://dientuvietnam.net/cgi-bin/mimetex.cgi?2\geq{z}) (3)
Từ 1 2 3 => http://dientuvietnam.net/cgi-bin/mimetex.cgi?P\leq{\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{z}{x}}
Do http://dientuvietnam.net/cgi-bin/mimetex.cgi?y\leq{z}\leq{2}\leq{2y}=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{2y}{z}\geq{1};\dfrac{z}{y}\geq{1}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?(\dfrac{2y}{z}-1)(\dfrac{z}{y}-\dfrac{1}{2})\geq{0}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{y}{z}+\dfrac{z}{y}\leq{\dfrac{5}{2}}
Tương tự : http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{x}{z}+\dfrac{z}{x}\leq{\dfrac{5}{2}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?P\leq{5} "=" khi x=1,y=1,z=2;
P/s : sao cái lệnh private ko xài đc nhỉ ?
Tôi nghĩ đề là :
Tìm max :
http://dientuvietnam.net/cgi-bin/mimetex.cgi?P=\dfrac{x^3+y^3+z^3}{xyz}
biết http://dientuvietnam.net/cgi-bin/mimetex.cgi?P=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}
G/s http://dientuvietnam...imetex.cgi?(x-y)(x^2-y^2)\geq{0}
<=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?xy^2+yz^2-xz^2\geq{y^3}
<= http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{y}{z}+\dfrac{z}{x}-\dfrac{z}{y}\geq{\dfrac{x^2}{yz}} (1)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{x^2}{yz}\leq{\dfrac{x}{z}} (do http://dientuvietnam.net/cgi-bin/mimetex.cgi?1\leq{x}\leq{y} ) (2)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{z^2}{xy}\leq{\dfrac{2z}{y}} (do http://dientuvietnam.net/cgi-bin/mimetex.cgi?1\leq{x} và http://dientuvietnam.net/cgi-bin/mimetex.cgi?2\geq{z}) (3)
Từ 1 2 3 => http://dientuvietnam.net/cgi-bin/mimetex.cgi?P\leq{\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{z}{x}}
Do http://dientuvietnam.net/cgi-bin/mimetex.cgi?y\leq{z}\leq{2}\leq{2y}=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{2y}{z}\geq{1};\dfrac{z}{y}\geq{1}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?(\dfrac{2y}{z}-1)(\dfrac{z}{y}-\dfrac{1}{2})\geq{0}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{y}{z}+\dfrac{z}{y}\leq{\dfrac{5}{2}}
Tương tự : http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{x}{z}+\dfrac{z}{x}\leq{\dfrac{5}{2}
=> http://dientuvietnam.net/cgi-bin/mimetex.cgi?P\leq{5} "=" khi x=1,y=1,z=2;
P/s : sao cái lệnh private ko xài đc nhỉ ?
Bài viết đã được chỉnh sửa nội dung bởi ocking: 30-10-2005 - 17:55
#5
Đã gửi 18-12-2005 - 09:20
Không phải là thế này: a+b+c=4 và a,b,c 1
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