已知x,y>0,求证:根号下(x^2+y^2)+2/(1/x+1/y)>(x+y)/2+根号下(xy)

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已知x,y>0,求证:根号下(x^2+y^2)+2/(1/x+1/y)>(x+y)/2+根号下(xy)
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已知x,y>0,求证:根号下(x^2+y^2)+2/(1/x+1/y)>(x+y)/2+根号下(xy)
已知x,y>0,求证:根号下(x^2+y^2)+2/(1/x+1/y)>(x+y)/2+根号下(xy)

已知x,y>0,求证:根号下(x^2+y^2)+2/(1/x+1/y)>(x+y)/2+根号下(xy)
√[(x²+y²)/2]+2/(1/x+1/y)≥(x+y)/2+√﹙xy﹚
√[(x²+y²)/2]-√﹙xy﹚≥(x+y)/2-2/(1/x+1/y)
﹛√[(x²+y²)/2]﹜²-[√﹙xy﹚]²/﹛√[(x²+y²)/2]+√﹙xy﹚﹜≥(x+y)/2-2xy/(x+y)
[﹙x-y)²/2]/﹛√[(x²+y²)/2]+√﹙xy﹚﹜≥﹙x-y)²/[2﹙x+y﹚]
√[(x²+y²)/2]+√﹙xy﹚≤x+y(当x≠y时﹚
(x²+y²)/2+xy+2√[(x²+y²)/2]√﹙xy﹚≤x²+y²+2xy
x²+y²+2xy+4√[(x²+y²)/2]√﹙xy﹚≤2x²+2y²+4xy
2√(x²+y²)√﹙2xy﹚≤x²+y²+2xy
此式显然成立,且以上步步可逆
∴√(x²+y²)+2/(1/x+1/y)>√[(x²+y²)/2]+2/(1/x+1/y)≥(x+y)/2+√﹙xy﹚(当x≠y时﹚
当x=y时,√(x²+y²)+2/(1/x+1/y)=√2x+x,
(x+y)/2+√﹙xy﹚=x+x
∴√(x²+y²)+2/(1/x+1/y)>(x+y)/2+√﹙xy﹚
故√(x²+y²)+2/(1/x+1/y)>(x+y)/2+√﹙xy﹚