If $S$ and $T$ are subgroups of a finite group $G$, prove that $|S||T|\leq |S \cap T||\langle S, T\rangle|$.
My approach is: since $\langle S, T\rangle$ is the smallest subgroup containg $S$ and $T$, it should at least contain all the elements of the form $st$ where $s$ belongs to $S$ and $t$ belongs to $T$.
Since $S$ has $|S|$ elements, $T$ has $|T|$ elements, $|\langle S, T\rangle| \geq |S||T|$.
Since $S$ and $T$ are both subgroups of $G$, they should at least both contain $1$ (unit). Therefore $|S \cap T|\geq 1$.
Putting them together, $|S||T|\leq |S \cap T||\langle S, T\rangle|$.
Is my proof right? Thanks