设∑ab=k
∑a/(a^2+2bc) ≤1=∑a/k
∑(a/k-a/(a^2+2bc))≥0
局部通分后发现是Schur型
∑a(a-b)(a-c)/k/(a^2+2bc)≥0
∑a(a-b)(a-c)/(a^2+2bc)≥0
因全对称性设a>=b>=c
∑a(a-b)(a-c)/(a^2+2bc)
≥a(a-b)(a-c)/(a^2+2bc)+b(b-c)(b-a)/(b^2+2ca)
=(a-b)^2c(2a^2+3ab+2b^2-2ac-2bc)/(a^2+2bc)/(b^2+2ca)
≥0
取等条件同Schur