If a > b > C is known, it is proved by analysis or synthesis that 1 / (a + b) + 1 / (B-C) > = 4 / (A-C)

If a > b > C is known, it is proved by analysis or synthesis that 1 / (a + b) + 1 / (B-C) > = 4 / (A-C)

Because a-c + A-B = B-C, and a > b > C
So A-B = B-C > = 2 √ (a-b) * (B-C)
So (1 / (a-b)) * (1 / (B-C)) > = 4 / (a-b) ^ 2
Because 1 / (a-b) + 1 / (B-C) > = 2 √ 1 / ((a-b) * (B-C))
So 1 / (a-b) + 1 / (B-C) > = 2 √ 4 / (a-b) ^ 2
So 1 / (a + b) + 1 / (B-C) > = 4 / (A-C)
How about it? Do you understand? We need to use twice mean inequality