- 证明题已知x、y、z>0,且x^2/(1+x^2)+y^2
- 已知x、y、z>0,且x^2/(1+x^2)+y^2/(1+y^2)+z^2/(1+z^2)＝2.
求证：x/(1+x^2)+y/(1+y^2)+z/(1+z^2)≤√2。
- 条件式即1/(1+x^2)+1/(1+y^2)+1/(1+z^2)＝1,
设a＝1/(1+x^2),b＝1/(1+y^2),c＝1/(1+z^2),
则a+b+c＝1,且以所设代入原式得
x/(1+x^2)+y/(1+y^2)+z/(1+z^2)
＝√(1-a)·√a+√(1-b)·√b+√(1x-c)·√c
≤√(1-a+1-b+1-c)·√(a+b+c)
＝√2,
故原不等式得证。