Let f (x) = ax ^ 2 + BX + C, and f (1) = - A / 2,3a > 2C > 2B, prove a > 0 and - 3
F (1) = a + B + C = - A / 2, that is 3A + 2B + 2C = 0, and 3a > 2C > 2B, obviously there is a > 0 (if A2C > 2B, 3A > - 3a-2b > 2B can be obtained, for the first unequal sign, - 3 can be obtained
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