代数很菜,大概是做麻烦了(以下sigma视为对i从1到n求和)
原不等式等价于sigma(bi^2)sigma(ai^3/ai+bi)≥sigma(aibi)sigma(ai^2bi/ai+bi)
等价于sigma(bi(ai+bi))sigma(ai^3/ai+bi)≥sigma(aibi)sigma(ai^2)
由cauchy不等式sigma(ai^3/ai+bi)≥(sigma(ai^2))^2/sigma((ai+bi)ai)
只需证明sigma(bi(ai+bi))sigma(ai^2)≥sigma(aibi)sigma((ai+bi)ai)
也即sigma(ai^2)sigma(bi^2)≥(sigma(aibi))^2
此即为cauchy不等式
原不等式等价于sigma(bi^2)sigma(ai^3/ai+bi)≥sigma(aibi)sigma(ai^2bi/ai+bi)
等价于sigma(bi(ai+bi))sigma(ai^3/ai+bi)≥sigma(aibi)sigma(ai^2)
由cauchy不等式sigma(ai^3/ai+bi)≥(sigma(ai^2))^2/sigma((ai+bi)ai)
只需证明sigma(bi(ai+bi))sigma(ai^2)≥sigma(aibi)sigma((ai+bi)ai)
也即sigma(ai^2)sigma(bi^2)≥(sigma(aibi))^2
此即为cauchy不等式