设abc是三角形的三边，S是三角形的面积，求证：c^2-a^2-b^2+4ab>=(4√3)S

那是4根号3喔

由余弦定理有：

c^=a^+b^-2abcosC

c^2-a^2-b^2+4ab

=2ab(2-cosC)

而S=1/2absinC

(4√3)S=2√3absinC

c^2-a^2-b^2+4ab-(4√3)S

=2ab(2-cosC)-2√3absinC

=2ab[2-(cosC+√3sinC)]

=2ab[2-2sin(C+30)]

因为2sin(C+30)≤2

2-2sin(C+30)≥0，2ab>0

故c^2-a^2-b^2+4ab-(4√3)S≥0

即c^2-a^2-b^2+4abS≥（4√3)S

证明: a^2+b^2+c^2-4√3S [S=(1/2)absinC] =a^2+b^2+c^2-(4√3)(1/2)absinC =a^2+b^2+(a^2+b^2-2abcosC)-(2√3)absinC =2<a^2+b^2-2ab{(1/2)cosC+[(√3)/2]sinC}> =2[a^2+b^2-2absin(C+30度)] 因为sin(C+30度)≤1 所以:2absin(C+30度)≥-2ab 所以:2[a^2+b^2-2absin(C+30度)]≥2(a^2+b^2-2ab) =2(a-b)^2≥0 所以:c^2-a^2-b^2+4ab≥(4√3)S

由余弦定理：c^2-a^2-b^2=-2abcosC S=(1/2)absinC 不等式变为:-2abcosC+4ab>=2√3absinC 既证明2√3absinC+2abcosC<=4ab √3sinC+cosC<=2 而√3sinC+cosC=2(sinCcos30+cosCsin30)=2sin(C+30)<=2 故命题得证