证明 配方, 得
2[∑(a + b + c + d)(b + c + d + e)(c + d + e + f) - 48(a + d)(b + e)(c + f)]
=(a+b+d+e)[(a+b-c-f)²+(c-d-e+f)²]
+(b+c+e+f)[(b+c-d-a)²+(d-e-f+a)²]
+(c+d+f+a)[(c+d-e-b)²+(e-f-a+b)²]
+(a-b+d-e)²(a+b+10c+d+e+10f)
+(b-c+e-f)²(b+c+10d+e+f+10a)
+(c-d+f-a)²(c+d+10e+f+a+10b)
≥0.
