Let $0<p<1$ snd $\dfrac{1}{p}+\dfrac{1}{p'}=1$. Notice that $p'<0$. If $f \in L^{p}$ and $0<\int_{\Omega}\vert g(x) \vert^{p'}dx < \infty$ then, $$\int_{\Omega}\vert f(x)g(x) \vert dx \geq (\int_{\Omega}\vert f(x) \vert^{p}dx)^{\frac{1}{p}}(\int_{\Omega}\vert g(x) \vert^{p'}dx)^{\frac{1}{p'}} $$
The assumption on $g$ implies that $g>0$ almost surely.
Write $h(x) = f(x)g(x)$. Without loss of generality (by multiplying $g$ and $h$ by constants) we can assume $$ \left(\int |g|^{p'} dx \right)^{\frac{1}{p'}}= 1 = \int |h| dx $$ So what we want to prove reduces to $$ \int |h / g|^p \leq 1 $$
To prove this, apply the regular Holder inequality: put $h^p$ in $L^{1/p}$ and $1/g^p$ in $L^{1/1-p}$. Observe that
$$ \frac{1}{|g|^{p/(1-p)}} = |g|^{p'}$$