a>0,b>0,c>0,求证1/a+1/b+1/c>=2*[1/(a+b)+1/(b+c)+1/(c+a)]

a>0,b>0,c>0,求证1/a+1/b+1/c>=2*[1/(a+b)+1/(b+c)+1/(c+a)]
a>0,b>0,c>0,求证1/a+1/b+1/c>=2*[1/(a+b)+1/(b+c)+1/(c+a)]

a>0,b>0,c>0,求证1/a+1/b+1/c>=2*[1/(a+b)+1/(b+c)+1/(c+a)]
1/a+1/b-4/(a+b) =[b(a+b)+a(a+b)-4ab)/[ab(a+b)] =(a-b)^2/[ab(a+b)] >=0当a=b等号成立 所以：1/a+1/b>=4/(a+b) 同理1/a+1/c>=4/(a+c),1/b+1/c>=4/(b+c) 相加：2(1/a+1/b+1/c)>=4[1/(a+b)+1/(b+c)+1/(a+c)] 所以：1/a+1/b+1/c>=2[1/(a+b)+1/(b+c)+1/(a+c)]