a+2b+3c=2*[(a+2b+3c)/2]=(1/2)(1/a+2/b+3/c)(a+2b+3c)=(1/2)(1+2b/a+3c/a+2a/b+4+6c/b+3a/c+6b/c+9)=(1/2)[(2b/a+2a/b)+(3c/a+3a/c)+(6c/b+6b/c)+14]由均值不等式,有：原式>=(1/2)[2√(2b/a)(2a/b) +2√(3c/a)(3a/c...