Divergence of integral I found in web the following problem:

Let $f,g$ be continuous non-negative decreasing functions on $\mathbb R_{\ge 0}$ such that $$\int_0^\infty f(x)\,dx\to\infty, \int_0^\infty g(x)\,dx\to\infty$$. Define the function $h$ by $h(x)=\min \{f(x),g(x)\},\forall x\in\mathbb R_{\ge 0}$. Prove or disprove that $$\int_0^\infty h(x)\,dx$$ also diverges.

I bet the statement is true, so I attempt to prove it. Firstly, since $h$ is also continuous, non-negative and decreasing, we know that the statement is equivalent to say

Let $f,g$ be continuous non-negative decreasing functions defined on $\mathbb N$, show that
  $$(\sum_{k=0}^\infty f(k)\to\infty\land \sum_{k=0}^\infty g(x)\to\infty )\implies \sum_{k=0}^\infty \min\{f(k),g(k)\}\to\infty $$

Now, if one has assumed the limits
$$\lim_{n\to\infty} f(n)/\min\{f(n),g(n)\}, \lim_{n\to\infty} \min\{f(n),g(n)\}/g(n), \lim_{n\to\infty} f(n)/g(n)$$ all exist, I think I could give a proof.
Let $$\lim_{n\to\infty} f(n)/\min\{f(n),g(n)\}=L_1, \lim_{n\to\infty} \min\{f(n),g(n)\}/g(n)=L_2, \lim_{n\to\infty} f(n)/g(n)=L$$. We know that any of $L_1,L_2, L$ is nonnegative. Since $L_1L_2=L$, if $L>0$, then we are done by the limit comparison test. If $L=0$, then in the long run $\min\{f(x),g(x)\}\sim f(x)$, we are done as well.
However, I could hardly think of an approach if the assumption is not made. Any help would be appreciated. Thanks in advance.

here is my attempt to prove (it's only a rought idea)
let define
$$M(x)=max(f,g)$$
and
$$I_1(x)=\frac{M(x)+h(x)}{2}$$
thus by comparison $I_1(x)$ diverges
then define
$$I_k(x)=\frac{I_{k-1}(x)+h(x)}{2}$$
thus by induction $I_k(x)$ diverges
as $$I_k(x) \rightarrow h(x)$$
$h(x)$ also diverges $\square$?
