Description: 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
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暴力破解不等式.pas
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