ABC为三角形,求证：[tan(A/2)]^2+[tan(B/2)]^2+[tan(C/2)]^2>=1

∵（a-b）^2≥0,∴（a^2+b^2）/2≥ab.当且仅当a=b时,“=”成立.
同理,(b-c)^2≥0,(c-a)^2≥0,∴（b^2+c^2）/2≥bc,（c^2+a^2）/2≥ca
当且仅当b=c,c=a时,“=”成立.
三式相加,得：a^2+b^2+c^2≥ab+bc+ca,当且仅当a=b=c时,“=”成立.
即,当a=b=c时a^2+b^2+c^2最小.
此题,当[tan(A/2)]=[tan(B/2)]=[tan(C/2)]时
[tan(A/2)]^2+[tan(B/2)]^2+[tan(C/2)]^2最小.
即A/2=B/2=C/2=π/3时原式最小.
而当A/2=B/2=C/2=π/3时,[tan(A/2)]^2+[tan(B/2)]^2+[tan(C/2)]^2=1
∴[tan(A/2)]^2+[tan(B/2)]^2+[tan(C/2)]^2>=1