证明: 配方, 得
64[∑(1-a)²∏(bc+a)²-∑a²(1-b²)(1-c²)(ca+b)²(ab+c)²]
=∑a(b-c)²[(1-2a)²{56a²bc+a[a²(32c²+15bc+32b²)+17abc(b+c)+98b²c²]
+abc[28a^3+4a²(b+c)+8a(8b²+bc+8c²)+45bc(b+c)]
+4b²c²[19a^3+13bc(b+c)]+16b^3c^3(a(b+c)+b²+bc+c²)}
+abc{8a+13bc+19bc(b+c)+12bc(b²+c²)
+4bc[a(7b²+26bc+7c²)+(b+c)(12b²+25bc+12c²)]
+bc[32a²(b²+c²)+16(b²+3bc+c²)(a(b+c)+2b²+bc+2c²)]}]≥0.
64[∑(1-a)²∏(bc+a)²-∑a²(1-b²)(1-c²)(ca+b)²(ab+c)²]
=∑a(b-c)²[(1-2a)²{56a²bc+a[a²(32c²+15bc+32b²)+17abc(b+c)+98b²c²]
+abc[28a^3+4a²(b+c)+8a(8b²+bc+8c²)+45bc(b+c)]
+4b²c²[19a^3+13bc(b+c)]+16b^3c^3(a(b+c)+b²+bc+c²)}
+abc{8a+13bc+19bc(b+c)+12bc(b²+c²)
+4bc[a(7b²+26bc+7c²)+(b+c)(12b²+25bc+12c²)]
+bc[32a²(b²+c²)+16(b²+3bc+c²)(a(b+c)+2b²+bc+2c²)]}]≥0.
