Prove by analysis: a square b square + b square C square + C square a square ≥ ABC (a + B + C) I just asked. Unfortunately, I read the answer. Here is the analytical method, not the difference method, Using basic inequalities and so on

Prove by analysis: a square b square + b square C square + C square a square ≥ ABC (a + B + C) I just asked. Unfortunately, I read the answer. Here is the analytical method, not the difference method, Using basic inequalities and so on

The proof of the basic inequality of mathematics published by Jiangsu Education Press
(1) Find the maximum value of lgx + logx10 (x > 1), and find the value of X at the maximum value. (the second X is the base of logarithmic function)
(2) If the above question is changed to 0 < x < 1, what will be the result?

lgX+logx10
=lgX+1/lgX
Because x > 1, lgx > 0,1 / lgx > 0
Therefore, according to a + B ≥ 2 √ AB, we get the following results:
lgX+1/lgX≥2
So: it has a minimum value of 2. When lgx = 1 / lgx, the minimum value is obtained, and then x = 10
(2) When 0 < x < 1, lgx < 0, 1 / lgx < 0
（－lgX）+（－1/lgX）≥2
lgX+1/lgX=－[（－lgX）+（－1/lgX）]≤－2
When x = 1 / 10, the maximum value is - 2

The proof of inequality in the first grade of Higher Education
Let ab ≠ 0, using the basic inequality, we have the following proof: B / A + A / b = (B2 + A2) / AB ≥ 2Ab / AB = 2, try to judge whether the proof process is correct, if correct, please explain the basis of each step; if not, please explain the reason

B / A + A / b = (B2 + A2) / ab
≥ 2Ab / AB... Incorrect if ab