using covering space technique,prove that if $G$ is a group with subgroups $H$ and $K$ then $$[G:H \cap K] \leq [G:H][G:K]$$
I couldn't understand the relation between them and the covering space,so it means that I am missing something.so it will be great if you give me some hints,thank you very much.
This is not a complete answer but it's too long to be a comment.
I would guess that the relation has something to do with the fact that every group is realizable as the fundamental group of some connected CW-complex of dimension $\geq 2$. So you would realize $G$ as $\pi_1(X)$ and then there would exist covering spaces of this $X$ with fundamental groups equal to $H,K,$ and $H\cap K$. There is a correspondence $[\pi_1(X):\pi_1(\tilde{X})]=$ degree of the covering $\tilde{X}\rightarrow X$ that holds whenever a space and its covering are path connected. However connected CW-complexes are always path-connected, and the coverings that correspond to the subgroups of $\pi_1(X)$ are all connected CW-complexes, just like $X$.
The argument proceeds by drawing a diagram and applying the unique lifting property to lift a covering of $X$ with fundamental group $H\cap K$ into coverings of $X$ with fundamental groups $H$ and $K$. Using this you can get a commutative diagram which should allow you to prove the result based on arguments about the degrees of the coverings.