若导数y=f(x)在R上可导且满足不等式xf'(x)>-f(x)恒成立,且常数a,b满足a>b,求证:af(a)>bf(b).
问题描述:
若导数y=f(x)在R上可导且满足不等式xf'(x)>-f(x)恒成立,且常数a,b满足a>b,求证:af(a)>bf(b).
答
xf'(x)>-f(x)
xf'(x)+f(x)>0
[xf(x)]′>0
xf(x)在R上是单调递增函数,且常数a,b满足a>b,
故af(a)>bf(b).