Question

定义在(0,正无穷)的函数f(x)满足f(x)+f(y)=f(xy)且x>1时f(x)<0

定义在(0,正无穷)的函数f(x)满足f(x)+f(y)=f(xy)且x>1时f(x)<0若不等式f(根号下x^2+y^2)<f(根号下xy)+f(a)对任意xy属于（0，正无穷）恒成立则a范围

Answer 1

f(x)+f(x)=f(x^2)，所以f(x)=f(x^2)/2
f[√(x^2+y^2)]<f[√(xy)]+f(a)
f(x^2+y^2)/2<f(xy)/2+f(a)
f(x^2+y^2)<f(xy)+2f(a)
f(x^2+y^2)<f(xy)+f(a^2)

又f(x)+f(1/x)=f(1)
f(1)+f(1)=f(1)
所以f(1/x)=-f(x)

所以f(a^2)=-f(1/a^2)
f(xy)=-f[1/(xy)]

所以f(x^2+y^2)<f(xy)+f(a^2)
f(x^2+y^2)<-f[1/(xy)]-f(1/a^2)
f(x^2+y^2)+f[1/(xy)]+f(1/a^2)<0
f[(x^2+y^2)/(xya^2)]<0

又x>1时f(x)<0
所以(x^2+y^2)/(xya^2)>1恒成立
即(x^2+y^2)/(xy)>a^2
只要(x^2+y^2)/(xy)的最小值比a^2大，就恒成立
x^2+y^2≥2xy
(x^2+y^2)/(xy)≥2
所以2>a^2时，恒成立
-√2<a<√2

Answer 2

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