f(x)在R上函数,且对于任意ab∈R.满足f(ab)=af(b)+bf(a) 当X>1时,f(x)恒正,若a>b>0 求证:bf(a)>af(b)
问题描述:
f(x)在R上函数,且对于任意ab∈R.满足f(ab)=af(b)+bf(a) 当X>1时,f(x)恒正,若a>b>0 求证:bf(a)>af(b)
答
当a,b不为0时,f(ab)=af(b)+bf(a)可化为f(ab)/(ab)=f(a)/a+f(b)/b令g(x)=f(x)/x,则g(ab)=g(a)+g(b),且x>1时,g(x)=f(x)/x>0设00,从而g(x)在R+上是增函数.从而,若a>b>0,则g(a)>g(b),即f(a)/a>f(b)/b,bf(a)>af(b)...