It is known that the differentiable function f (x) defined in (0, + ∞) satisfies XF '(x) - f (x) > 0 and f (x) > 0 (1) Let f (x) = f (x) / x, and prove that f (x) is an increasing function on (0, positive infinity) (2) If a > b > 0, AF (a) and BF (b) were compared

It is known that the differentiable function f (x) defined in (0, + ∞) satisfies XF '(x) - f (x) > 0 and f (x) > 0 (1) Let f (x) = f (x) / x, and prove that f (x) is an increasing function on (0, positive infinity) (2) If a > b > 0, AF (a) and BF (b) were compared

(1)F(x)=f(x)/x
F'(x)=[xf'(x)-f(x)]/x^2
Because XF '(x) - f (x) > 0, so
F'(x)>0
thus
F (x) is an increasing function on (0, positive infinity)
(2) Let g (x) = XF (x)
G '(x) = f (x) + XF' (x) > 2F (x) > 0 (because XF '(x) - f (x) > 0, f (x) > 0)
So g (x) = XF (x) is an increasing function on (0, positive infinity)
a>b>0,af(a)>bf(b)