{Abs[b(c-b)(d-c)] > Abs[(b+c)^2 d], 0<b, 0<c, 0<d, d>(b+c+d)/3}{Abs[b(c-b)(d-c)] > Abs[(b+c)^2 d], 0<b, 0<c, 0<d, d>(b+c+d)/3}

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