Cho: 13<x12y1.Tìm GTNN của: P=x2+y2+x2y2[(4x1)yx]2
Xét f(x)=x2+y2+x2y2(4xyxy)2,x(13,12]
Ta có: f(x)=2x+2y2x(4xyxy)22x2y2(4xyxy)(4y1)(4xyxy)4 
                          =2x+2xy3(4xyxy)3 
                          =2x[(4xyxy)3y3](4xyxy)3 
Mà ta có: 4xyx2y=2y(2x1)x2(2x1)x=3x20, vì x12 
                   4xyxy=y(4x1)x(4x1)x=3x1>0, vì x>13
Vậy f(x)0,x(13,12] .
Suy ra: Pf(12)=14+y2+y2(2y1)2 
Xét: g(y)=14+y2+y2(2y1)2,y1 
Ta có: g(y)=4y(y1)(4y22y+1)(2y1)30,y1 
Suy ra: Pg(1)=94 
Vậy MinP=94x=12,y=1 
Thế thì phải đợi học đạo hàm bạn nhé. –  khangnguyenthanh 07-10-12 12:12 AM
Tại em chưa học nên không biết ạ :D –  Xusint 07-10-12 12:09 AM
Nếu chưa học đạo hàm thì mình làm thế nào anh Khánh nhỉ? –  Xusint 06-10-12 11:50 PM
áp dụng BĐT Cauchy :
x2y2[(4x1)yx]2+(4x1)yxxy+(4x1)yxxy3
x2y2[(4x1)yx]23-2[(4x1)yx]xy=-5+2x+2y(1)
x2+18x+18x34x234-14x(2)
y2+1y+1y3y23-2y(3)
cộng từng vế của (1)(2)(3)P54+74x
do x1274x72
P94

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