Proving the divergence of a infinite integral

Could anyone give me some ideas/tips or solutions, how to get this task done? It's my homework in real analysis and I'm quite confused about it.

The task is as follows:

Let functions $f,g: [a,\infty) \to \mathbb{R}$ be of such kind, that $f(x) \geq 0$ and $g(x) > 0$, when $x \in [a, \infty)$.
Let $\int_a^\infty f(x) dx$ be a divergent integral. Prove that at least one of the integrals

$$ \int_a^\infty f(x)g(x)dx, \qquad \int_a^\infty\frac{f(x)}{g(x)} dx$$

diverges.

Solution 1:

By Holder inequality \begin{align*} \int_{a}^{\infty}f(x)dx&=\int_{a}^{\infty}\left(\frac{f(x)}{g(x)}\right)^{1/2}(f(x)g(x))^{1/2}dx\\ &\leq\left(\int_{a}^{\infty}\frac{f(x)}{g(x)}dx\right)^{1/2}\left(\int_{a}^{\infty}f(x)g(x)dx\right)^{1/2}\\ &<\infty, \end{align*} if the both integrals are convergent.

Solution 2:

If $a\ge0$ one can use elementary methods:

Assume that both $\int_{a}^{\infty}f(x)g(x)dx$ and $\int_{a}^{\infty}\dfrac{f(x)}{g(x)}dx$ converge.

then $$\int_{a}^{\infty}f(x)g(x)+\dfrac{f(x)}{g(x)}dx$$ converge.

We can see that:

$$f(x)g(x)+\dfrac{f(x)}{g(x)}=\dfrac{f(x)g^{2}(x)+f(x)}{g(x)}=\dfrac{f(x)(g^{2}(x)+1)}{g(x)}=f(x)\cdot\dfrac{g^{2}(x)+1}{g(x)}$$

Therefore :

$$\int_{a}^{\infty}f(x)g(x)+\dfrac{f(x)}{g(x)}dx=\int_{a}^{\infty}f(x)\cdot\dfrac{g^{2}(x)+1}{g(x)}dx$$

Since $2\le\dfrac{x^{2}+1}{x}$ for every $x>0$ then $2f(x)<f(x)\cdot\dfrac{g^{2}(x)+1}{g(x)}$ for

every x>0 and by the monotonicity of the integral we get:

$$\int_{a}^{\infty}2f(x)dx\le\int_{a}^{\infty}f(x)\cdot\dfrac{g^{2}(x)+1}{g(x)}dx$$

Then:

$$0\le\int_{a}^{\infty}f(x)dx\le2\int_{a}^{\infty}f(x)dx\le\int_{a}^{\infty}f(x)g(x)+\dfrac{f(x)}{g(x)}dx$$

But $\int_{a}^{\infty}f(x)dx$ diverges, which is a contradiction.

So atleast one of the integrals $\int_{a}^{\infty}f(x)g(x)dx$ , $\int_{a}^{\infty}\dfrac{f(x)}{g(x)}dx $ diverge.