Let T be a tree with t edges and G a graph. Prove that if |E(G)| ≥ t · |V (G)|, then T is a subgraph of G. I tried to prove it by induction on t. Obviously true when t=1 and 2. Suppose true when t=k, then when tree T has t=k+1, we could remove a leaf x from T and thus go back to case t=k, but I have no idea how to extend that to a tree isomorphic to T.
There are similar questions which restrict on degree of vertices of G, I also tried to use $$\sum_{v\in G}{d(v)}=2|E(G)|$$ but did not find a way. (e.g. Trees that are isomorphic to a subgraph of a graph G.)
 A: Assume the statement for some $k\geqslant1$. Let $G$ be a graph with $|E|\geqslant (k+1)|V|$ and $T$ a tree with $k+1$ edges. The degree of any vertex in $G$ is at most $|V|-1$, so if $G'$ is a subgraph of $G$ obtained by removing a vertex, 
$$|E'|\geqslant (k+1)|V|-(|V|-1)=k|V|+1\geqslant k(|V|-1). $$
Therefore if $T'$ is a subgraph of $T$ obtained by removing an edge, there is a subgraph $S'$ of $G'$ with $S'\cong T'$. Since the vertex removed from $G$ was arbitrary, it follows that there is a subgraph of $G$ congruent to $T$.
A: We prove by induction on $t$. (Assume G is not a null graph)
We can easily verify that when t=1,2 the statement is true. 
Let $T$ be a tree with t $edges$, then it is easy to see that $\max\limits_{v \in T} deg(v) = t$ 
Suppose true when t-1, then when t:
We have $$|E(G)| \geq t |V(G)|$$
Let U be the set of vertices of with degree not greater than $t-1$ in G
We have $$|E(G)|\leq|E(G\setminus U)|+(t-1)|U|$$
$$|V(G)|=|V(G\setminus U)|+|U|$$ 
Combining them, we get $$|E(G\backslash U)| \geq t |V(G\backslash U)|+|U| \geq(t-1)|V(G\backslash U)|$$
Which means we can embed $T'$ in $G\backslash U$ (where $T' :=T-x$, $x$ is a leaf of $T$),   now we can add back that leaf in $G$ since all vertices in $G\backslash U$ has degree larger than t (considering its degree in $G$), but in $T'$ maximum degree is $t-1$.
