判别式法证明不等式a,b,c≥0a, b, c \ge0a,b,c≥0, 求证:8(b2−ac)(c2−ab)9a(bc)(b−c)2≥08(b^2 - ac)(c^2 - ab) 9a(b c)(b - c)^2 \ge 08(b2−ac)(c2−ab)9a(bc)(b−c)2≥0.改编自《数学不等式第一卷》[罗马尼亚] Vasile2.30证明:8(b2−ac)(c2−ab)8(b2c2−ab3−ac3a2bc)8(b^2 - ac)(c^2 - ab) 8(b^2c^2 - ab^3 - ac^3 a^2bc)8(b2−ac)(c2−ab)8(b2c2−ab3−ac3a2bc)9a(bc)(b−c)29a(b3−b2c−bc2c3)9a(b c)(b - c)^2 9a(b^3 - b^2c - bc^2 c^3)9a(bc)(b−c)29a(b3−b2c−bc2c3)将aaa视为变量. 只需证明8bca2(bc)(b2−10bcc2)a8b2c2≥08bca^2 (bc)(b^2-10bcc^2)a 8b^2c^2\ge 08bca2(bc)(b2−10bcc2)a8b2c2≥0这等价于判别式小于等于0. 计算判别式, 得Δ−(b−c)2[16bc(b−c)2(b2−6bcc2)2]≤0\Delta - (b - c)^2 \left[ 16bc(b - c)^2 (b^2 - 6bc c^2)^2 \right] \le 0Δ−(b−c)2[16bc(b−c)2(b2−6bcc2)2]≤0
