已知a>0,b>0,c>0,a+b+c=1,求证1/a+1/b+1/c≥9(a+b+c)(1/a+1/b+1/c)=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1+(b+c)/a+1+(a+c)/b+1+(a+b)/c=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)≥3+2+2+2=91/a+1/b+1/c≥9