已知a>b>0,求证(a-b)^2/8a<(a+b)/2-SQR(ab)<(a-b)^2/8b

原式可转为
（a-b)²/8a<(√b-√a)²/2<（a-b)²/8b
因a>b>0,设y=√b，x=√a(x>y)
<=>（y²-x²）/2x<y-x<(y²-x²)/2y
证明：（y²-x²）/2x<y-x
<=>x+y<2x
<=>y<x，故不等式左边成立
证明：y-x<(y²-x²)/2y
<=>2y<x+y
<=>y<x故不等式右边也成立