Specifically, assume that conclusion B holds, and then take the conclusion as a condition to see what results can be deduced in reverse. Suppose that conclusion C can be deduced from conclusion B (that is, B =>C), and then check whether B and C are reversible (that is, whether C =>b), and if so, it is BC. Then analyze what you can get from C, if you can get the CD, continue to wait and so on ... BCD.H
When we find from A= > that H is easy to prove, then there is A = & gtHB.
So you can get a =>b. The original proposition is proved.
For example:
Let a and b be positive real numbers, 2c >;; A+B. Verification: under the symbol of C root (C 2-AB)
This problem is difficult to prove if it starts from known conditions. Consider using backward reasoning.
Prove: (C 2-AB) under C- radical sign < a < under c+ radical sign (c 2-ab)
-under the root sign (c 2-ab)
Absolute value (a-c)