The gist of this is that, given $f, g \in J$, where J contains all Riemann integrable functions on a compact rectangle, then $\sqrt{\int_K (f+g)^2}\leq \sqrt{\int_Kf^2} + \sqrt{\int_K g^2}$.
Now, I tried removing the radicals, resulting in ${\int_K (f+g)^2}\leq {\int_Kf^2} + {\int_K g^2} + 2{\int_Kf^2} {\int_K g^2} \iff {\int_K f^2+\int_K g^2 +2\int_K fg}\leq {\int_Kf^2} + {\int_K g^2} + 2\sqrt{\int_Kf^2} \sqrt{\int_K g^2}$.
I want to think there's a way to show that $\int_K fg\leq \sqrt{\int_Kf^2} \sqrt{\int_K g^2}$ in order to end the proof, but I can't think of any.
I'm barely beginning to understand compact sets and integration in compact sets, so any advice would be appreciated.
Cheers!