I can prove only that it's true for $e>400$. Edited: Actually it must be $n>400$ where $n$ is the no. of vertices.

Nice. Can you show that though?

Can anyone give a complete proof on this problem? I've tried to solve it but failed.

The problem should be “Given that there exists an Eulerian cycle” Also hint:(there is no $K_{3,3}$)

Here is a possible solution $v$ is the number of vertices and $e$ is the number of edges Let's count the number of beautiful pairs $(a,\{x,y,z\})$ where $\{a,x\}, \{a,y\}, \{a,z\} \in E(G)$ The key observation is that there is no $K_{3,3}$ so there exists at most 2 $a$ such that $(a,\{x,y,z\})$ is a beautiful pair by Jensen's inequality ${{\frac{2e}{v}}\choose{3}}v \le \Sigma_{i=1}^{v} {{d_i}\choose{3}} =(beautiful pairs) \le 2{{v}\choose{3}}$ this is equivalent to $v^5-3v^4+2v^3-2ev^2+6e^2v-4e^3 \ge0$ and this yields $\frac{e}{v}\le 0.3+{(\frac{e}{2})}^{5/2}$