x>y>0,xy=1,求证:(x^2+y^2)/x-y)>=2倍根号2
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x>y>0,xy=1,求证:(x^2+y^2)/x-y)>=2倍根号2
![x>y>0,xy=1,求证:(x^2+y^2)/x-y)>=2倍根号2](/uploads/image/z/4207034-2-4.jpg?t=x%3Ey%3E0%2Cxy%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%88x%5E2%2By%5E2%EF%BC%89%2Fx-y%EF%BC%89%3E%3D2%E5%80%8D%E6%A0%B9%E5%8F%B72)
(x^2+y^2)/x-y)
=(x^2+y^2-2xy+2xy)/(x-y)
=(x-y)^2/(x-y)+2xy/(x-y)
=(x-y)+2xy/(x-y)
因为均值不等式a+b>=2√ab
所以上式>=2√2xy (xy=1)
即 :(x^2+y^2)/(x-y)>=2√2
=(x^2+y^2-2xy+2xy)/(x-y)
=(x-y)^2/(x-y)+2xy/(x-y)
=(x-y)+2xy/(x-y)
因为均值不等式a+b>=2√ab
所以上式>=2√2xy (xy=1)
即 :(x^2+y^2)/(x-y)>=2√2
x>y>0,xy=1,求证:(x^2+y^2)/x-y)>=2倍根号2
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