Question

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

Answer 1

解答：
因为a^2=b^2+c^2-2bccosA
S=(1/2)bcsinA
则a^2+b^2+c^2-4√3S
=b^2+c^2-2bccosA+b^2+c^2-4√3*(1/2)bcsinA
=2b^2+2c^2-2bccosA-2√3bcsinA
=2b^2+2c^2-4bc[(1/2)cosA+(√3/2)sinA]
=2b^2+2c^2-4bc+4bc-4bccos(60-A)
=2(b-c)^2+4bc[1-cos(60-A)]

-120 < 60-A < 60
-1/2 < cos(60-A) ≤ 1
0 ≤ 1-cos(60-A) < 3/2
所以
a^2+b^2+c^2-4√3S = 2(b-c)^2+4bc[1-cos(60-A)] ≥0
当b=c且A=60时，即等边三角形时，等号成立

Answer 2

因为S=1/2ab*sinc
即证：c^2 -a^2 -b^2 + 4ab>=2ab√3sinc
即(b^2+a^2-c^2)/(2ab)<=2-√3sinc
即cosc+√3sinc<=2
即sin(c+30)<=1
显然成立

Answer 3

证明：根据余弦定理，可得：c² - a² -b² = -2ab *cosC
∴左边 = 2ab(2 - cosC)
∵S = 1/2 ab*sinC
∴令M = 2ab （2 - cosC）- 2√3 ab*sinC
=4ab （1 - 1/2 *cosC - √3/2* sinC)
∵1/2 *cosC + √3/2* sinC
=sin π/6*cosC + cosπ/6*sinC
=sin(C+π/6) ≤1
∴M≥0
故：结论成立