Inequality with sums and products

Question

Let a,b,c,d positive real numbers such that $a+b<c+d$ and $(a+b)(c+d)<ab+cd$. Prove that $(a+b)cd>(c+d)ab$

Source: book on olympiads I tried manipulating the given statements and i obtained $3ab<cd$ To use this i would need to show that $cd<3(a+b)$ but i am stuck.

Answer 1

Since $ab+cd>(a+b)(c+d)$ and $ab<a(a+b)<a(c+d)$, $cd>b(c+d)$. Thus $(a+b)cd>acd>ab(c+d)$.