若三角形ABC的三边分别为a,b,c,面积为S,求证：a^2+b^2+c^2大于等于4根号3S

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) ...