设x2>x1>0,则x2/x1>1
=>f(x2/x1)>0
f(x2)=f(x1*x2/x1)=f(x1)+f(x2/x1)>f(x1)
因此f(x)在0到无穷单调递增(i)
f(x)=f(3x*1/3)=f(3x)+f(1/3)=f(3x)-1
=>f(3x)-f(x)=1
=>f(9x)-f(3x)=1
=>f(9x)-f(x)=2 (ii)
f（x）-f（1/（x-2））≥2 (iii)
综合（i)(ii)(iii)=>
x>9/(x-2)
=>x-9/(x-2)>=0
=>(x^2-2x-9)(x-2)>=0
=>x>=1+√10 （0