If , and are distinct positive numbers, then the expression is
- A.positive
- B.negative
- C.non-positive
- D.non-negative
Show answer and solution
Correct option: B
Solution
Let x = b+c-a, y = c+a-b, z = a+b-c. Note x+y+z = a+b+c > 0, and x+y = 2c > 0, y+z = 2a > 0, x+z = 2b > 0, so x, y, z can be positive or negative, but their sum is positive.
Use AM-GM: for any positive reals, (y+z)(z+x)(x+y) ≥ 8xyz, i.e., 2a·2b·2c ≥ 8xyz, giving abc ≥ xyz = (b+c-a)(c+a-b)(a+b-c).
So (b+c-a)(c+a-b)(a+b-c) - abc ≤ 0, with equality only when x=y=z, i.e., a=b=c. Since a, b, c are distinct, the inequality is strict: the expression is strictly negative.
Use AM-GM: for any positive reals, (y+z)(z+x)(x+y) ≥ 8xyz, i.e., 2a·2b·2c ≥ 8xyz, giving abc ≥ xyz = (b+c-a)(c+a-b)(a+b-c).
So (b+c-a)(c+a-b)(a+b-c) - abc ≤ 0, with equality only when x=y=z, i.e., a=b=c. Since a, b, c are distinct, the inequality is strict: the expression is strictly negative.
[1991]