M=ba^2+cb^2+ac^2 = b*a*a+c*b*b+a*c*c
N=ab^2+bc^2+ca^2 = a*b*b+b*c*c+c*a*a
1.已知a>b>c,假定a,b,c都為正數(shù)
設(shè)a=3,b=2,c=1
M=ba^2+cb^2+ac^2 = 2*9+1*4+3*1 = 18+4+3 = 25
N=ab^2+bc^2+ca^2 = 3*4+2*1+1*9 = 12+2+9 = 23
M > N
2.假定a,b,c都為負數(shù)
設(shè)a=-1,b=-2,c=-3
M=ba^2+cb^2+ac^2 = (-2)*1+(-3)*4+(-1)*9 = -2-12-9 = -23
N=ab^2+bc^2+ca^2 = (-1)*4+(-2)*9+(-3)*1 = -4-18-3 = -25
M>N
綜合上情況M>N