How to prove that (A's square + 1) times (B's square + 1) times (C's square + 1) is greater than or equal to 8abc

How to prove that (A's square + 1) times (B's square + 1) times (C's square + 1) is greater than or equal to 8abc

There is a formula: a + b > = 2 (AB) ^ (1 / 2), which means that the sum of a + B is greater than or equal to 2 times the product of a times B
This formula can be extended to infinite polynomials, such as a + B + C > = 3 (ABC) ^ (1 / 3), the product of a times b times C under the sum of a + B + C is greater than or equal to 3 times the third root
So the original problem: (a ^ 2 + 1) (b ^ 2 + 1) (C ^ 2 + 1) is expanded into (ABC) ^ 2 + (AB) ^ 2 + (AC) ^ 2 + (BC) ^ 2 + A ^ 2 + B ^ 2 + C ^ 2 + 1, which is the sum of eight numbers. With the above formula, the octave power of ABC under the octave root sign should be greater than or equal to eight times, which is 8abc
So we have to prove it
I haven't done high school math problems for a long time. It's really time-consuming. Add points. Make sure it's right. It's really hard to use the computer to type formulas