很多不等式在展开以后形成如下的对称形式
sigma(s1^a1*s2^a2*...*sn^an)>=sigma(s1^b1*s2^b2*...*sn^bn)
（当然 作为齐次不等式 a1+a2+....an=b1+b2+...bn 变量s1,s2,...sn非负）
其中sigma表示对称和（也就是说 一共n!项） 例如
sigma(x^3)=x^3y^0z^0+x^3z^0y^0+y^3x^0z^0+y^3z^0x^0+z^3x^0y^0+z^3y^0x^0=2*(x^3+y^3+z^3)
sigma(x^3y^2z^1)=x^3y^2z^1+x^3z^2y^1+y^3x^2z^1+y^3z^2x^1+z^3x^2y^1+z^3y^2x^1
（三元sigma 一共是6项）
有时候 我们把sigma(s1^a1*s2^a2*...*sn*an)写作 [a1,a2,...an]
例如 著名的均值不等式可以写成 [n,0,0...0]>=[1,1,1...1]
又比如x^2+y^2+z^2>=xy+yz+zx 写成[2,0]>=[1,1]
本程序能比较两个完全对称不等式的大小关系。
After a lot of inequality start to form in the following symmetric form of sigma (s1 ^ a1* s2 ^ a2*...* sn ^ an)> = sigma (s1 ^ b1* s2 ^ b2*...* sn ^ bn) (of course, homogeneous inequality as a1+ a2+.... an = b1+ b2+ ... bn variables s1, s2, ... sn non-negative) that the symmetric and one sigma (ie a total of n! items) such as sigma (x ^ 3) = x ^ 3y ^ 0z ^ 0+ x ^ 3z ^ 0y ^ 0+ y ^ 3x ^ 0z ^ 0+ y ^ 3z ^ 0x ^ 0+ z ^ 3x ^ 0y ^ 0+ z ^ 3y ^ 0x ^ 0 = 2* (x ^ 3+ y ^ 3+ z ^ 3) sigma (x ^ 3y ^ 2z ^ 1) = x ^ 3y ^ 2z ^ 1+ x ^ 3z ^ 2y ^ 1+ y ^ 3x ^ 2z ^ 1+ y ^ 3z ^ 2x ^ 1+ z ^ 3x ^ 2y ^ 1+ z ^ 3y ^ 2x ^ 1 (three sigma is a total of six) at times we have sigma (s1 ^ a1* s2 ^ a2*.. .* sn* an) Writing [a1, a2, ... an] such as the well-known inequality can be written in the mean [n, 0,0 ... 0]> = [1,1,1 ... 1] and For example, x ^ 2+ y ^ 2+ z ^ 2> = xy+ yz+ zx written [2,0]> = [1,1] This procedure can compare two completely symmetrical relationship between the si (2009-05-22, Pascal, 1KB, 下载2次)