Question

一道不等式题急！！

设集合A由适合以下性质的函数f(x)构成：对任意x>0,y>0,且x≠y,都有f(x)+2f(y)>3f[(x+2y)/3],(1)试判断f1(x)=log2(x)及f2(x)=(x+1)^2是否在集合A中说明理由

Answer 1

log2(x)+2log2(y)=log2(xy²)

3log2((x+2y)/3)=log2((x+2y)/3)³

log2(x)是单调递增函数,只需证明

xy²>((x+2y)/3)³

只要举出反例便可证明f1(x)不在A中：

令x=6 y=3(取足够大即可)

左边=36 右边=4^3=64 左边<右边 矛盾

所以f1(x)不在集合A中

当f2(x)=(x+1)^2

f2(x)+2f2(y)=(x+1)²+2(y+1)²

3f2[(x+2y)/3]=3((x+2y)/3+1)²=(x+2y+3)²/3=((x+1)+2(y+2))²/3
=((x+1)²+4(x+1)(y+1)+4(y+1)²)/3

f2(x)+2f2(y)-3f2[(x+2y)/3]
=(3(x+1)²+6(y+1)²-(x+1)²-4(x+1)(y+1)-4(y+1)²)/3
=(2(x+1)²-4(x+1)(y+1)+2(y+1)²)/3
=2((x+1)²-2(x+1)(y+1)+(y+1)²)/3
=2((x+1)-(y+1))²/3
=2(x-y)²/3
因为x≠y,所以上式>0
即f2(x)+2f2(y)-3f2[(x+2y)/3]>0

即f2(x)+2f2(y)>3f2[(x+2y)/3]

所以f2(x)在集合A中.