Then (A 2+B 2+C 2) 2 > = 9 * (ABC) (4/3)

That is to say, only 9 * (ABC) (4/3) > = 3 (a 3b+b 3c+c 3a) is needed.

That is 3 * (ABC) (4/3) > =1(a 3b+b 3c+c 3a)

Variant 3> = (a 3b+b 3c+c 3a)/(ABC) (4/3)

Just prove that the above formula is true.

If there is a basic inequality, it is easy to get the above formula, so the original formula holds.