I'm trying to prove Holder's inequality using that in a measure space $(X,\mu)$ for every $h:X\to \mathbb{C}$ measurable, $0<\theta<1$ and $0<p<r<q<\infty$, with $$\frac{1}{p}(1-\theta)+\frac{1}{q}\theta=\frac{1}{r},$$ we have $$\|h\|_{r}\leq \|h\|_{p}^{1-\theta}\|h\|_{q}^{\theta}.$$ It is an exercise (probably simple) from a Tao's blog post.
He claims that we can choose $r$, $\theta$, $h$ and $\mu$ proving the Holder's inequality.
The Holder's inequality statement I'm trying to prove is the following:
Let $(Y,\nu)$ be a measure space. Prove that for every measurable functions $f,g:Y \to \mathbb{C}$ and $p',q'>0$ such that $$\frac{1}{p'}+\frac{1}{q'}=1$$ we have $$\|fg\|_1 \leq \|f\|_{p'}\|g\|_{q'}.$$
I tryied using $\theta=\frac{1}{2}$ and $r=2$ so that $\frac{1}{p}+\frac{1}{q}=1$. Then trying to fit $c:=\frac{d\mu}{d\nu}$ and $h$ such that \begin{cases} h^2 c = f g \\ h^p c = f^p \\ h^q c = g^q \end{cases} but it doesn't seem to work.
I appreciate any hints or solutions!
Edit 1: I also tryied only fixing $\theta=1/2$. That is, $$\frac{2}{r}=\frac{1}{p}+\frac{1}{q}.$$ Then we would like to have $p'=\frac{2p}{r}$ and $q'=\frac{2q}{r}$ and \begin{cases} h^r c = f g \\ h^p c = f^{p'} \\ h^q c = g^{q'} \end{cases} I couldn't see a way of solving it without using $\frac{1}{f}$ or $\frac{1}{g}$. But we don't know if they are not null or even integrable.