Prove the Inequality – $\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right)$

Question

Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=2$. Prove the inequality:
$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right)$
I came across this problem in an online lecture, the presenter gave us this problem as an exercise to do at home. I haven't been able to solve it, but u have some ideas I don't think work.
First, I thought about distributing the $4$ across the fractions, so we would get
$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4=\frac{a^2+c^2+2ac+2ad+2bc}{ad+bc}+\frac{b^2+d^{2}+2bd+2ac+2bd}{ac+bd}$
The thing is, I got two $2bd$ in the second fraction, so it isn't kind of cyclic like I thought it would be, or at least similar to the first fraction. I have been stuck on this problem since then and don't know how to continue, maybe I am wrong.

Answer 1

By CS inequality, $$\frac{(a+c)^2}{ad+bc}+\frac{(b+d)^2}{ac+bd}+\frac{2^2}1\geqslant \frac{(a+c+b+d+2)^2}{(a+b)(c+d)+1}$$

It remains to show that for $a+b=x\in (0, 2)$, $$\frac{16}{x(2-x)+1} \geqslant 4\left(\frac{x+1}{3-x} +\frac{3-x}{x+1}\right)$$ $$\iff 4(3-x)(x+1)\geqslant (x(2-x)+1)((x+1)^2+(3-x)^2)$$ $$\iff 2(x-1)^4\geqslant 0$$