已知x>y>0,xy=1,求证(x^2+y^2)/(x-y)≥2根号2
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已知x>y>0,xy=1,求证(x^2+y^2)/(x-y)≥2根号2
![已知x>y>0,xy=1,求证(x^2+y^2)/(x-y)≥2根号2](/uploads/image/z/6444045-45-5.jpg?t=%E5%B7%B2%E7%9F%A5x%3Ey%3E0%2Cxy%3D1%2C%E6%B1%82%E8%AF%81%28x%5E2%2By%5E2%29%2F%28x-y%29%E2%89%A52%E6%A0%B9%E5%8F%B72)
已知x>y>0,xy=1
(x^2+y^2)/(x-y)
=(x^2-2+y^2+2)/(x-y)
=(x^2-2xy+y^2+2)/(x-y)
=[(x-y)^2+2]/(x-y)=x-y+2/(x-y)≥2根号2
(x^2+y^2)/(x-y)
=(x^2-2+y^2+2)/(x-y)
=(x^2-2xy+y^2+2)/(x-y)
=[(x-y)^2+2]/(x-y)=x-y+2/(x-y)≥2根号2
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