证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc已知a,b,c是正实数,证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc②已知ABC中A,B,C所对的边分别为A,B,C,三角形的面积为S求证:C^2-A^2-B^2+4AB大于等
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![证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc已知a,b,c是正实数,证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc②已知ABC中A,B,C所对的边分别为A,B,C,三角形的面积为S求证:C^2-A^2-B^2+4AB大于等](/uploads/image/z/110954-2-4.jpg?t=%E8%AF%81%E6%98%8E%3A1%2F%28a3%2Bb3%2Babc%29%2B1%2F%28b3%2Bc3%2Babc%29%2B1%2F%28c3%2Ba3%2Babc%29%E2%89%A41%2Fabc%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%98%AF%E6%AD%A3%E5%AE%9E%E6%95%B0%2C%E8%AF%81%E6%98%8E%EF%BC%9A1%2F%EF%BC%88a3%2Bb3%2Babc%29%2B1%2F%28b3%2Bc3%2Babc%29%2B1%2F%28c3%2Ba3%2Babc%29%E2%89%A41%2Fabc%E2%91%A1%E5%B7%B2%E7%9F%A5ABC%E4%B8%ADA%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAA%2CB%2CC%2C%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BAS%E6%B1%82%E8%AF%81%EF%BC%9AC%EF%BC%BE2-A%EF%BC%BE2-B%EF%BC%BE2%2B4AB%E5%A4%A7%E4%BA%8E%E7%AD%89)
证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc已知a,b,c是正实数,证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc②已知ABC中A,B,C所对的边分别为A,B,C,三角形的面积为S求证:C^2-A^2-B^2+4AB大于等
证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc
已知a,b,c是正实数,证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc
②已知ABC中A,B,C所对的边分别为A,B,C,三角形的面积为S
求证:C^2-A^2-B^2+4AB大于等于四倍根号三S
第一题中的3是3次方
证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc已知a,b,c是正实数,证明:1/(a3+b3+abc)+1/(b3+c3+abc)+1/(c3+a3+abc)≤1/abc②已知ABC中A,B,C所对的边分别为A,B,C,三角形的面积为S求证:C^2-A^2-B^2+4AB大于等
【解】去分母并化简,原式等价于
a6(b3+c3)+b6(c3+a3)+c6(a3+b3)≥2a2b2c2(a3+b3+c3)
(1)
由对称性,不妨设a≥b≥c.
因为
2a2b2c2(a3+b3+c3)≤(a4+b4)c4+(b4+c4)a5+(c4+a4)b5
而
a6(b3+c3)+b6(c3+a3)+c6(a3+b3)-(a4+b4)c5-(b4+c4)a5-(c4+a4)b5
=a5b3(a-b)+a5c3(a-c)-b5a3(a-b)+b5c3(b-c)-c5a3(a-c)-c5b3(b-c)
=(a-b)a3b3(a2-b2)+(a-c)a3c3(a2-c2)+(b-c)b3c3(b2-c2)≥0
所以(1)成立.
第2题化为三角函数做
三元一次不等式而已,对于本天才来说实在是太简单了!
先列成方程组,尽可能化简.
先用加减消元法,中途用代入,结尾用回加减.
基本上可以知道a.b.c的取值范围,
结果就一步一步推出.