由a+b+c=1和均值定理得1+a=a+b+c+a≥4(a^2bc)^(1/4)1+b=a+b+c+b≥4(ab^2c)^(1/4)1+b=a+b+c+c≥4(abc^2)^(1/4)上面3式相乘得(1+a)(1+b)(1+c)≥64abc两边同除abc得(1+1/a)(1+1/b)(1+1/c)≥64