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x>y>0,xy=1,求证:(x^2+y^2)/x-y)>=2倍根号2
人气:259 ℃ 时间:2020-06-21 06:24:29
解答
(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
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