{Abs[b(c-b)(d-c)] > Abs[(b+d/2)(c-b)(c+d/2)], 0<b, 0<c, 0<d, d<c}{Abs[b(c-b)(d-c)] > Abs[(b+d/2)(c-b)(c+d/2)], 0<b, 0<c, 0<d, d<c}

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