令 g(x) = f(x)/x
g'(x) = d(f(x)/x)/dx = [xf'(x)- f(x)]/x^2
因xf'(x) > f(x), => xf'(x) - f(x) > 0 => 当x 不为 0 时,g'(x) > 0
因此g(x)在x不为零时是增函数 => 当a>1时,g(a) > g(1),即f(a)/a > f(1)/1 => f(a) > af(1).