令x=ab,y=ac,z=bc,则xyz=1不妨设x≥y≥z,则x+y≥x+z≥y+z∴1/(y+z)≥1/(x+z)≥1/(x+y)由顺序和≥乱序和,得x/(y+z)+y/(x+z)+z/(x+y)≥y/(y+z)+z/(x+z)+x/(x+y)x/(y+z)+y/(x+z)+z/(x+y)≥z/(y+z)+x/(x+z)+y/(x+y)上面...