How to prove $1 \leq ||f||^p_p ||g||^p_p $?

Question

Given $\mu(X)=1$, and suppose $p\in (0,\infty)$, and let $f,g$ be positive measurable functions on $X$ such that $fg \geq 1$. How to prove

$$1\leq \int_X f^p d\mu \cdot \int_X g^p d\mu?$$

Since $f$ and $g$ are non-negative and $fg\geq 1$, we have $f>0$ and $g>0$. Then apply Holder's inequality to get

$$||fg||_1\leq ||f||_p ||g||_q$$

where $p$ and $q$ are conjugate exponents.

We can see that $||f||_p\leq||f||_p^p$ if $p\geq1$. Then, I tried to prove $$||g||_p^p \geq ||g||_q = ||g||_{\frac{p}{p-1}}.$$

How can I proceed to prove $||g||_p^p \geq ||g||_q$?

Or there are other ways to prove $$1\leq \int_X f^p d\mu \cdot \int_X g^p d\mu?$$

Answer 1

$\|fg\|_{p/2}\leq\|f\|_{p}\|g\|_{p}$, but $|fg|\geq 1$, so $|fg|^{p/2}\geq 1$. Since $\mu(X)=1$, hence $\|fg\|_{p/2}\geq 1$, so $\|f\|_{p}\|g\|_{p}\geq 1$, finally, $\|f\|_{p}^{p}\|g\|_{p}^{p}\geq 1$.