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)

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)
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

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)
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

∵x²/(1+x²)+y²/(1+y²)+z²/(1+z²) = 2,
∴1/(1+x²)+1/(1+y²)+1/(1+z²) = (1-x²/(1+x²))+(1-y²/(1+y²))+(1-z²/(1+z²)) = 1.
∵2√2...

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∵x²/(1+x²)+y²/(1+y²)+z²/(1+z²) = 2,
∴1/(1+x²)+1/(1+y²)+1/(1+z²) = (1-x²/(1+x²))+(1-y²/(1+y²))+(1-z²/(1+z²)) = 1.
∵2√2·x ≤ x²+2 (∵(x-√2)² ≥ 0),
∴2√2·x/(1+x²) ≤ x²/(1+x²)+2/(1+x²).
同理2√2·y/(1+y²) ≤ y²/(1+y²)+2/(1+y²), 2√2·z/(1+z²) ≤ z²/(1+z²)+2/(1+z²).
相加得2√2·(x/(1+x²)+y/(1+y²)+z/(1+z²))
≤ (x²/(1+x²)+y²/(1+y²)+z²/(1+z²))+2(1/(1+x²)+1/(1+y²)+1/(1+z²))
= 4.
即x/(1+x²)+y/(1+y²)+z/(1+z²) ≤ √2.
注: 上述证明并不需要x, y, z > 0的条件.

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