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+
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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.
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