Proving the number of leaves is larger by at least two than the number of vertices with a degree of at least 3 
Prove that in every tree, the number of leaves is larger by at least two than the number of vertices with a degree of at least 3.

Trying induction, I get something that is too short to be right, and I don't see how to use $\sum d(v)=2(n-1)$ because we have to prove here that $\displaystyle\sum_{d(v)=1}1 \ge \sum_{d(v)\ge 3}1+2 $ and there's no accounting here for the $d(v)=2$.
Another thing I noticed is that every vertex in a tree such that $d(v) > 1$ leads to an amount $> 1$ of leaves. So for every single $d(v)=3$ there will be at least two leaves. But can I prove that? any hints?

Induction is the way, but in a more simple way:
If you remove a leaf from the tree, it is still a tree, and you have three cases
1 - the father of the leaf you removed is now a leaf, so the number of leaves and nodes with degree $\ge$3 are both the same
2 - the father of the leaf you removed had exactly degree 3, so now it has degree 2, and the number of leaves and nodes with degree $\ge$3 both decreased by one
3 - the father of the leaf you removed had degree $
>$3, so now it has degree $\ge$3, and the number of leaves has decreased by one, and nodes with degree $\ge$3  remained the same
In each case, the induction works
