x>0,y>0,x+y=1,求证log2(x^2·y^2+1)-log2x-log2y≥log2(17)-2

x>0,y>0,x+y=1,求证log2(x^2·y^2+1)-log2x-log2y≥log2(17)-2
x>0,y>0,x+y=1,求证log2(x^2·y^2+1)-log2x-log2y≥log2(17)-2

x>0,y>0,x+y=1,求证log2(x^2·y^2+1)-log2x-log2y≥log2(17)-2
把式子的左边化为log2 (x^2*y^2+1)-log2 (xy)=log2 (x^2*y^2+1)/xy=log2 xy+(1/xy)
由x+y=1得x^2+y^2+2xy=1
因为x^2+y^2≥2xy
所以2xy+2xy≥1
可解得xy≥1/4
所以xy+(1/xy)≥17/4
即log2(x^2·y^2+1)-log2x-log2y≥log2 17/4=log2(17)-2