Given that ABC is a real number with complementary equality, we prove that a ^ 4 + B ^ 4 + C ^ 4 > ABC (a + B + C)

Given that ABC is a real number with complementary equality, we prove that a ^ 4 + B ^ 4 + C ^ 4 > ABC (a + B + C)

Let x = a ^ 2, y = B ^ 2, z = C ^ 2
a^4+b^4+c^4
=x^2+y^2+z^2
=1/2((x^2+y^2)+(x^2+z^2)+(y^2+z^2))
>=XY + XZ + YZ (x ^ 2, y ^ 2, Z ^ 2 are all greater than or equal to zero)
=a^2b^2+b^2c^2+c^2a^2
=1/2((a^2b^2+b^2c^2)+(b^2c^2+c^2a^2)+(a^2b^2+c^2a^2))
>=B ^ 2Ac + C ^ 2Ab + A ^ 2BC (a ^ 2B ^ 2, B ^ 2C ^ 2, a ^ 2C ^ 2 are greater than or equal to 0)
=abc(a+b+c)
∵ if a = b = C, ABC is not equal to each other
therefore
a^4+b^4+c^4>abc(a+b+c)