$\forall v (d_v \ge (\delta \ge 4))\Rightarrow $ $G$ has two cycles of length at least $\delta+1$ If every degree's vertex of graph $G(V,E)$ bigger than $\delta\ge 4$. Prove $G$ has two cycles of length at-least  $\delta+1$ with disjoint edges.
 A: I have two ideas to reach the above result, I hope that someone can develop them.
$1^{st}$ Idea:
By probabilistic method: 
number of edges in $G$ graph is at least $$\frac{\delta v(G)}{2}\geq 2v(G)$$
So by choosing at least $\frac{1}{2}\delta v(G)$ edges from $\frac{1}{2}v(G)(v(G)-1)$ we must be sure that we have two cycles that are edge disjoint of lenght at least $\delta +1$ each.
$2^{nd}$ Idea:
By taking the subgraph $T'_{\delta,\delta}$ ( which is a  sort of tree subgraph as described below).
A $d$-regular tree is a tree in which all vertices have a degree of $d$ or 1 ( degree of leaves is 1, and degree of the  rest vertices is $d$). 
Let $T_{d,l}$ be $d$-regular tree with $l$ levels. 
Example: the following graph is $T_{3,4}$

Let $T'_{d,l}$ be the graph defined from $T_{d,l}$ in the following way:  connect  each leave( i.e. vertex of degree one) to  exactly one other leave from every branch.
Notes: 1-branch is the set of vertices that all have the same neigbor of the root as there ancestor.
2-ancestors of a vertex are the set of vertices that form the path from the root of the tree to this vertex.
Example: The following graph is $T'_{3,3}$

Now by taking the subgraph $T'_{\delta,\delta}$  we must reach the result.
