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6.3 不等式的证明⑶

6.3 不等式的证明⑶. a > b a - b > 0 a = b a - b = 0 a < b a - b < 0. a > b > 1 a = b = 1 a < b < 1. a. a. a. b. b. b. 复习. ⒈ 比较法证明不等式的依据:. 作差法. 作商法 (a , b∈R + ). ⒉ 比较法证明不等式的步骤:. 判断符号 判断商与 1 的大小. 作差 作商. 变形.

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6.3 不等式的证明⑶

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  1. 6.3 不等式的证明⑶

  2. a>b a-b>0 a=b a-b=0 a<b a-b<0 a>b >1 a=b =1 a<b <1 a a a b b b 复习 ⒈ 比较法证明不等式的依据: 作差法 作商法(a,b∈R+) ⒉ 比较法证明不等式的步骤: 判断符号 判断商与1 的大小 作差 作商 变形

  3. 新课讲解 例⒈已知a,b,c是不全相等的正数, 求证:a(b2+c2)+b(c2+a2)+c(a2+b2)>6abc 证明:∵ b2+c2 ≥2bc,a>0 ∴ a(b2+c2)≥2abc. ① 同理 b(c2+a2)≥2abc. ② c(a2+b2)≥2abc. ③ ∵ a,b,c是不全相等的正数, ∴b2+c2≥2bc,c2+a2≥2ac,a2+b2≥2ab 三式不能全取“=”号 从而①、②、③三式不能全取“=”号。 ∴ a(b2+c2)+b(c2+a2)+c(a2+b2)>6abc

  4. 例⒉已知a,b,c为正数, 求证:a2b2+b2c2+c2a2≥abc(a+b+c) 证明: ∵a2b2+b2c2≥2ab2c ① b2c2+c2a2≥2bc2a ② a2b2+c2a2≥2ca2b ③ ∴由①+②+③得 2(a2b2+b2c2+c2a2)≥2abc(a+b+c) ∴ a2b2+b2c2+c2a2≥abc(a+b+c)

  5. ∴ lg +lg +lg >lga+lgb+lgc 求证:lg +lg +lg >lga+lgb+lgc ≥ ∴ ≥ ≥ a+b a+c a+b a+b a+b b+c b+c a+c b+c a+c b+c a+c 2 2 2 2 2 2 2 2 2 2 2 2 ac bc ab ∴ ··>abc 例⒊已知:a,b,c是不全相等的正数, 证明:∵ a,b,c是正数, 又∵a,b,c是不全相等的正数,

  6. 课堂小结 在不等式的证明中,利用某些已经证明过的不等式和不等式的性质推导出所要证明的不等式成立,这种证明方法叫做综合法。 课堂练习 若实数x与y满足x+y-4=0, 求x2+y2的最小值 。

  7. 课外作业 第15页  第1,2题 第18页  第6题

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