prove a Quad with the length of the sides being a,b,c and d, then its area S is such that: 4S < or =(a+c)(b+d)

Question

I need to prove that a Quadrilateral with the length of the sides being a,b,c and d, with an area of S, that 4S is less than or egual to (a+b)(c+d), somehow using twice the area of a triangle is absin Theta where theta is the angle between the sides of lengths a,b and sin theta is less than or egual to 1,with equality if theta is a right angle.

I need to prove that a Quadrilateral with the length of the sides being a,b,c and d, with an area of S, that 4S is less than or egual to (a+b)(c+d), somehow using twice the area of a triangle is absin Theta where theta is the angle between the sides of lengths a,b and sin theta is less than or egual to 1,with equality if theta is a right angle. The answer must have S in it! ie: S

Answer 1

If yo draw a diagonal you divide the quad into two triangles the area of the two triangles is 1/2 ab sin X + 1/2 cd sin Y.
If you draw the other diagonal, you will get the area of two more triangles that are 1/2 ad sin W +1/2 bcsin Z.
If you add thes together you wil have twice the area of the quadrilateral.

So, 2A = 1/2(absinX+cdsinY + adsinW + bcsinZ)
Now the greatest value the sin can have is 1 ,so
2A =< 1/2(ab + cd + ad + bc)
So 4A = 4S <= ab + cd +ad +bc)
4S <= a(b+d) + c(b+d) <= (a+b)(c+d)

Answer 2

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Answer 3

Well, I've got 4ab<=ac+ad+bc+bd
Also, i have 4ab=absin(theta)+absin(90-theta)
You were saying that sin(theta)<1, where 0 Substituting:
absin(theta)+bdsin(90-theta) Because sin(theta) + sin(90-theta)=1,
1
I'm confident you can take it from there.