A>=B>=C>=D>=E>=F>0
A+E+F=B+C+D
求證:BCD>=AEF

我想了很久又擱著一段時間在想仍解不出,雖然很不甘願,但是...還是忍痛求助...(有點討厭算不出來去問人的感覺說...感覺到自己的渺小...能力的不足@@")
拜託各位了...

令A=a1+F,B=a2+F,C=a3+F,D=a4+F,E=a5+F,則a1>=a2>=a3>=a4>=a5>=0
A+E+F=B+C+D-->a1+a5=a2+a3+a4
BCD= (a2+F)(a3+F)(a4+F)
=a2a3a4+(a2a3+a3a4+a4a2)*F+(a2+a3+a4) *F2+F3
AEF=a1a5F+(a1+a5)*F2+F3
BCD-AEF=a2a3a4+(a2a3+a3a4+a4a2)*F-a1a5*F
=a2a3a4+(a2a3+a3a4+a4a2-(a2+a3+a4-a5)a5)*F=a2a3a4+(a2a3+a3a4+a4a2-a2a5-a3a5-a4a5+a5*a5)*F
因a1>=a2>=a3>=a4>=a5>=0
故(a2a3+a3a4+a4a2-a2a5-a3a5-a4a5+a5*a5)>=0
所以BCD-AEF>=0,BCD>=AEF