If a>b, it can be concluded that AC 2 ≥ BC 2, the former is a sufficient condition for the latter;

If AC 2 ≥ BC 2 is known, a>b and the former are the necessary conditions for the latter.

If the above two points are satisfied at the same time, then A > B is a necessary and sufficient condition for AC 2 ≥ BC 2.

Prove:

First, the sufficient conditions are proved:

a & gtb

a-b & gt； 0

C≠0，c 2 > 0，a-b & gt； 0 c^2(a-b)>； 0 ac^2>； bc^2

When c=0, c 2 (a-b) = 0ac 2 = BC 2.

In a word, a>b, AC 2 ≥ BC 2.

Re-check the necessary conditions:

ac^2≥bc^2

c^2(a-b)≥0

C 2 is a constant, not negative. If the inequality holds, A-B≥0.

a≥b

AC 2 ≥ BC 2 can only deduce a≥b, but A >；; b

Therefore, this proposition is a false proposition, and a>b is a sufficient condition but not a necessary condition for AC 2 ≥ BC 2. When a=b, the inequality AC 2 ≥ BC 2 also holds. Therefore, a>b is not a necessary and sufficient condition for AC 2 ≥ BC 2.

It is correct to say that the proposition "a≥b is a necessary and sufficient condition for AC 2 ≥ BC 2".