If a, B and C belong to positive real numbers, prove that (a square + 1) (b square + 1) (C square + 1) is greater than or equal to 8abc

If a, B and C belong to positive real numbers, prove that (a square + 1) (b square + 1) (C square + 1) is greater than or equal to 8abc


a^2+1>=2a
b^2+1>=2b
c^2+1>=2c
So (a square + 1) (b square + 1) (C square + 1) is greater than or equal to 8abc



It is proved that if the square of a plus the square of B divides a by B, a equals B


There is something wrong with the expression. Maybe the square of a plus the square of B can be divided by ab
Let a square + b square = nab
So the square of a is B
The square of B | a
And a and B are integers, so | a | = | B |



It is known that A.B.C is a positive integer. What is the answer to prove that (a b) (b c) (c a) is greater than or equal to 8abc


It is known that A.B.C is a positive integer. What is the answer to prove that (a b) (b c) (c a) is greater than or equal to 8abc
(a b)(b c)(c a)=a²b²c²=(abc)²
(abc)²≥8abc
The answer is ABC ≥ 8