引用:
原帖由 chu1976 於 2008-5-27 09:06 PM 發表
三角形ABC中,求(1)(tanA/2)^2+(tanB/2)^2+(tanC/2)^2的最小值(2)(tanA/2)+(tanB/2)+(tanC/2)的最小值
因為 0<A, B, C<π,所以 tan(A/2)、tan(B/2)、tan(C/2)皆為正數。
(1)
利用柯西不等式,
{(tan(A/2))^2+(tan(B/2))^2+(tan(C/2))^2}
×{(tan(B/2))^2+(tan(C/2))^2+(tan(A/2))^2}
≧ {(tan(A/2))(tan(B/2))+(tan(B/2))(tan(C/2))+(tan(C/2))(tan(A/2))}^2=1^2
所以,(tan(A/2))^2+(tan(B/2))^2+(tan(C/2))^2 最小值為 1
(2)
{(tan(A/2))+(tan(B/2))+(tan(C/2))}^2
=(tan(A/2))^2+(tan(B/2))^2+(tan(C/2))^2
+2{(tan(A/2))(tan(B/2))+(tan(B/2))(tan(C/2))+(tan(C/2))(tan(A/2))}
≧1+2=3
所以,tan(A/2)+tan(B/2)+tan(C/2)≧√3