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设x,y均为正数,且x>y,求证:2x+1x2-2xy+y2≥2y+3.

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设x,y均为正数,且x>y,求证:2x+
1
x
设x,y均为正数,且x>y,求证:2x+1x2-2xy+y2≥2y+3.
证明:由题设x>y,可得x-y>0;
∵2x+
1
x2-2xy+y2-2y=2(x-y)+
1
(x-y)2=(x-y)+(x-y)+
1
(x-y)2;
又(x-y)+(x-y)+
1
(x-y)2≥3
3(x-y)2
1
(x-y)2
=3,当x-y=1时取“=“;
∴2x+
1
x2-2xy+y2-2y≥3,即2x+
1
x2-2xy+y2≥2y+3.