设a,b,c为正数,求证：(a^2+b^2)/2c+(b^2+c^2)/2a+(c^2+a^2)/2b

不妨设a≥b≥c＞0,则a^3≥b^3≥c^3,1/bc≥1/ac≥1/ab则左式为顺序和,即：a^3/bc+b^3/ca+c^3/ab≥a^2/c+b^2/a+c^2/b(乱序和)a^3/bc+b^3/ca+c^3/ab≥b^2/c+c^2/a+a^2/b(乱序和)两式相加,2(a^3/bc+b^3/ca+c^3/ab)≥(a^2...