Problem: Let $ P $ be a partition on a real interval; $P=\{a=x_0<x_1<\cdots<x_n=b\}$. Suppose we have two Riemann integrable functions denoted $f$ and $r$.
Then $$\sum^n_{i=0} \sup\lvert fr\rvert(x)(x_i-x_{i-1}) \le \sqrt{\sum^n_{i=0} \sup f^2(x)(x_i-x_{i-1})} \sqrt{\sum^n_{i=0} \sup r^2(x)(x_i-x_{i-1})}.$$
Attempt: I took the infimum of the equation over all partitions to get
$$\lvert \int fr \,dt \rvert \le \int \lvert fr \rvert \,dt \le \sqrt{\int f^2 \,dt} \sqrt{\int r^2 \,dt}$$
I know the Cauchy-Schwarz inequality is useful in this proof; which says
$$\lvert \sum^n_{i=1} a_jb_j \rvert ^2 \le \sum^n_{j=1} \lvert a_j \rvert ^2 \sum^n_{j=1} \lvert b_j \rvert ^2 $$
But I'm having trouble identifying my $a_j$ and $b_j$ in this case.
hint: $\text{sup}|f\cdot r|(x_i-x_{i-1}) \leq \text{sup}|f|\sqrt{(x_i-x_{i-1})}\cdot \text{sup}|r|\sqrt{(x_i-x_{i-1})}$.
$a_i = \text{sup}|f|\sqrt{(x_i-x_{i-1})}, b_i = \text{sup}|r|\sqrt{(x_i-x_{i-1})}$