分类：教育/科学-理工学科-数学
问题：角的不等式在ΔABC中，求证1/A^2+1/B^2+1/C^2≥ 
在ΔAB中，求证1/A^2+1/B^2+1/C^2≥27/π^2  (弧度)
回答：[π^2(1/A^2+1/B^2+1/C^2)]^(1/3)=
=[(A^(1/3))^3+(B^(1/3))^3+(C^(1/3)^3]^(1/3)*
*[(A^(1/3))^3+(B^(1/3))^3+(C^(1/3)^3]^(1/3)*
*[(1/A^(2/3))^3+(1/B^(2/3))^3+(1/C^(2/3)^3]^(1/3)≥
≥A^(1/3)A^(1/3)(1/A^(2/3)+B^(1/3)B^(1/3)(1/B^(2/3)+
+C^(1/3)C^(1/3)(1/C^(2/3)=3
==>
1/A^2+1/B^2+1/C^2≥27/π^2