<s>[INST] Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $ 
Let $x, y, z > 0$. Prove that
$$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$

I encountered this problem today and thought that it was straightforward, and to be honest it still might be I just can't figure it out for some reason.
I approached this problem using the famous Cauchy-Schwarz inequality. I applied the inequality to every denominator. For example for the first fraction:
$$(1+xy+xz)(1 + \frac{1}{x}+ \frac{1}{x}) \ge (1+y+z)^2$$
After doing this for every fraction I switched the denominator for every one of them with the left side of the Cauchy-Schwarz inequality. Reducing each fraction we are left with
$$\frac{1}{1+2/x} + \frac{1}{1+2/y} + \frac{1}{1+2/z} \ge 1 $$
I am not sure how to continue from here. I tried to use Titu's lemma but it didn't work. I will be very happy if someone can help me out here.
Edit: My Cauchy was wrong.Thanks to all who contributed :)
 [/INST] Hint :
Can you show ? For $a,b,c>0$
$$\frac{1+ba+bc}{(1+a+c)^{2}}-\frac{b^{2}+ba+bc}{(b+a+c)^{2}}=\frac{\left(b-1\right)^{2}\left(a+c\right)}{\left(a+c+1\right)^{2}\left(a+b+c\right)}\geq 0$$
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