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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√[(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﹚

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