题目是不是应该是三角形ABC的面积为(a^2+b^2+c^2)/4√3.
因为c^2=b^2+a^2-2abcosC
S=(1/2)absinC
则a^2+b^2+c^2-4√3S
=b^2+a^2-2abcosC+b^2+a^2-4√3*(1/2)absinC
=2b^2+2a^2-2abcosC-2√3absinC
=2b^2+2a^2-4ab[(1/2)cosC+(√3/2)sinC]
=2b^2+2a^2-4ab+4ab-4abcos(60-C)
=2(b-a)^2+4ab[1-cos(60-C)] =0
而(b-a)^2≥0,4ab[1-cos(60-C)] ≥0,所以只有各自等于0时才成立.
此时1-cos(60-C)=0,C=60°