Prove that the perimeter of any quadrilateral is greater than twice the length of any of its diagonal I am stuck with the following problem that says :

Prove that the perimeter of any quadrilateral is greater than twice the length of any of its diagonal.

My try: 
........
 
For any quadrilateral $ABCD\,,$ we can easily prove that 
$$AB+BC+CD+DA \gt AC+BD......\tag{1}$$ 
Now, three cases arise. Either, $$AC=BD\,\,or AC \gt BD\,\, or  AC \lt BD$$.
If $AC=BD$,then the result follows from (1). 
If $AC \gt BD \implies AC+BD \gt 2BD $  and then the result follows from (1).
If $ BD \gt AC \implies AC+BD \gt 2AC $  and then the result follows from (1).
Can someone take some time to check if I made any mistake or is there any better way to tackle the problem.
Thanks in advance for your time.
 A: Remove one of the diagonals to help the visual. Then you have two triangles. Apply triangle inequality.
A: $AB+AD>BD$ and $BC+CD>BD$, then 
$$AB+BC+CD+DA>2BD$$
A: It is just a straightforward implementation of the properties of a triangle.
$$AB + BC > AC$$
$$ AD + CD > AC $$
Similarly for the other diagonal.
A: We assume the original quadrilateral is nondegenerate (not a point or a line segment).
Pick a diagonal $D_1$, and consider the family of quadrilaterals obtained by moving the other two vertices linearly along the other diagonal $D_2$ toward the center. As the vertices move in, the perimeter of the resulting quadrilaterals strictly shrinks.
Initially, the perimeter is that of the original quadrilateral; in the limiting case, the vertices meet in the center and the perimeter has shrunk to exactly twice the length of $D_1$.
So the original perimeter is strictly less than twice the length of any given diagonal.
A: E be the meeting point of diagonals...
∆AEB , AE+EB > AB
∆EBC , EC + EB >BC
∆ CED, CE + DE > CD
∆ DEA, DE + EA > DA
Adding all above
2AC+ 2BD> AB+BC+CD+DA(Replacing BE and ED with BD, similarly with AC)
