Cho http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{a+b+c}{3}\le\dfrac{1}{4}\sqrt[3]{\dfrac{(b+c)^2(c+a)^2(a+b)^2}{abc}}
BDT
Started By euler, 01-11-2005 - 16:54
#1
Posted 01-11-2005 - 16:54
#2
Posted 01-11-2005 - 17:27
Đặt x=b+c, y=c+a, z=a+b, chuyển về 3 cạnh tam giác, ta có:
bdt
bdt
#3
Posted 01-11-2005 - 17:36
Có thể chứng minh bằng cách khác như sau:
Áp dụng bđt
và bđt
Áp dụng bđt
và bđt
you will never know what will you get untill you have really try.
from :...........................................................
from :...........................................................
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