An interesting math problem

Verify the regular solid shapes only can be five types.
证明正多面体只能有5种.
First: Look at the proof procedure below:

 1        /**//*证明正多面体只能有5种
 2         *分别是正 4， 6， 8， 12， 20面体
 3         *
 4         * First: 欧拉公式  V+F-E=2
 5         * 假设每个面均为 正m边形    每个顶点都有n条棱  >>  显然的 m>=3 & n>=3              1
 6         * Second: m*F = 2E
 7         * Third:  n*V = 2E
 8         * 
 9         * 由First, Second, Third得到下式
10         * 2E/n + 2E/m -E = 2  >>  1/n + 1/m = 1/E + 1/2
11         * 即 1/n + 1/m > 1/2
12         * 上式中得到 m & n 不能同时大于3                                                   2
13         * 
14         * Manual ：由1，2可得到 m，n至少一个等于3，分情况讨论后即可得出结果
15         * Automation ：1，2作为条件 得出结果 如下方法实现
16        */

Yes, it's just so easy :0
Then I try to carve it out with code
Second: The easiest solution

 1        static void GetResult()
 2        {
 3            for (double i = 3; i < int.MaxValue; i++)
 4            {
 5                for (double j = 3; j < int.MaxValue; j++)
 6                {
 7                    if (1/i + 1/j > 1.0/2.0)
 8                    {
 9                        Console.WriteLine("i={0},j={1}", i.ToString(),j.ToString());
10                    }
11                }
12            }
13        }

Also, it's very easy, and I'm sure it is not a good methd. The biggest problem is that the efficiency is so low.
It must has better solutions to solve this question.
Log it here and fix the efficiency problem later.