showing that $(\int f )(\int g) \geq 1$

Question

Let $\mu(X) =1$.
Let $f,g \in L^1(X)$ be two positive functions satisfying $f(x) g(x)>1$ for almost all $x$, Then $$\left(\int f ~dx\right) \left(\int g~dx\right) \geq 1.$$

Show also that if $f,g\in L^2(X)$ with $\int f ~dx= 0$, then $$\left(\int fg~dx\right)^2 \leq \left[ \int g^2 ~dx - \left(\int g~dx\right)^2 \right] \int f^2~dx.$$

I think I have to use Holder's inequality for both questions:

For the first question, since $\mu(X) =1$, $1\lt \int fg~dx$. How do I apply Holder's inequality.

Answer 1

For the first, try applying the inequality to $\int \sqrt{f g}$, and obtain a lower bound for $\int \sqrt{f g}$.

For the second, let $\overline{g} = \int g$, and apply the inequality to $\int f (g-\overline{g})$.

Answer 2

Hint:

For the first inequality use Hölder for $\sqrt{gf}$.