how to show that $ fgh \in L^1$ If $f \in L^p,$ $g \in L^q,$ $h \in L^r$ where $1/p +1/q +1/r=1$, then how to show $fgh \in L^1$ with $\int |fgh| \leq ||f||_{L^p} ||g||_{L^q} ||h||_{L^r}$
 A: The idea is as follows:
We know the inequality for two functions which is $$|fg|_1\le \|f\|_p\|g\|_q $$ where $\frac{1}{p}+\frac{1}{q}=1$. Here in this problem we have $$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1 $$ So, we can do like this
\begin{align*}
\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1\\
\implies \frac{1}{p}+\left(\frac{1}{q}+\frac{1}{r}\right)=1\\
\frac{1}{p}+\frac{1}{s}=1\ \ \text{where}\ s=\frac{pq}{p+q}
\end{align*}  So, we have $$\frac{1}{s}=\frac{1}{p}+\frac{1}{q}$$ or $$1=\frac{1}{p/s}+\frac{1}{q/s}.$$
         $$\int|fgh|\leq\|fg\|_{s}\|h\|_r\leq\|f\|_p\|g\|_q\|h\|_r$$  
First we show that $\|fg\|_{s}\leq \|f\|_p\|g\|_q$. This is easy since $$\|fg\|_{s}=\left(\int|fg|^{s}\right)^{\frac{1}{s}}\leq(\|f^{s}\|_{p/s}\|g^{s}\|_{q/s})^{\frac{1}{s}}=\|f\|_p\|g\|_q,$$ where we apply the Holder's inequality (it is permissible since $|f|\in L^p(\mathbb{R})$, thus $|f|^{s}\in L^{p/s}(\mathbb{R})$). As a result, $|fg|\in L^{s}(\mathbb{R})$. Apply Holder's inequality again, we get the first inequality in far above.
