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xyz都是正数,且x²/(1+x²)+y²/(1+y²)+z²/(1+

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xyz都是正数,且x²/(1+x²)+y²/(1+y²)+z²/(1+z²)=2,求证 x/(1+x²)+y/(1+y²)+z/(1+z²)<=根号2
x²/(1+x²)+y²/(1+y²)+z²/(1+z²)=2
则[1-1/(1+x²)]+[1-1/(1+y²)]+[1-1/(1+z²)]=2
所以1/(1+x²)+1/(1+y²)+1/(1+z²)=1
由柯西不等式
2=[x²/(1+x²)+y²/(1+y²)+z²/(1+z²)][1/(1+x²)+1/(1+y²)+1/(1+z²)]
≥[x/(1+x²)+y/(1+y²)+z/(1+z²)]²
所以x/(1+x²)+y/(1+y²)+z/(1+z²)≤√2