Prove that $\sum_{cyc}(1-x)^2\ge \sum_{cyc}\frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}$ 
Let $x,y,z>0$. Show that
$$\sum_{cyc}(1-x)^2\ge \sum_{cyc}\dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$

Source: In a blog entry posted in 2013, it was said that this problem was proposed by Dongyi Wei (full marks both at 50th IMO 2009 and 49th IMO 2008) and was solved by Zipei Nie (full marks at 51st IMO 2010).
Here's what I have done. The expression
$$\sum_{cyc}(1-x)\left((1-x)-\dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}\right)\ge 0$$
or
$$\sum_{cyc}(1-x)\cdot\dfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}\ge 0.$$
But this way does not help for the starting inequality.
 A: The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=\frac{1}{a}$
$y=\frac{1}{b}$
$z=\frac{1}{c}$
We get the following inequality :
$$\sum_{cyc}\frac{(1-a)^2}{a^2}\geq \sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[\frac{1}{4};\frac{1}{2}]$:
$$\frac{(1-a)^2}{a^2}\geq\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$\sum_{cyc}\frac{(1-a)^2}{a^2}\geq \sum_{cyc}\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6 \geq\sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[\frac{1}{4};\frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$\sum_{cyc}\frac{(1-x)^2}{x^2}\geq u\sum_{cyc}\frac{(1-a)^2}{a^2}\geq u\sum_{cyc}\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6\\ \geq u\sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}\geq \sum_{cyc}\frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=\cos^2 s$;
$y=\cos^2 v$;
$z=\cos^2 w$
And 
$a=\cos^2 p$;
$b=\cos^2 r$;
$c=\cos^2 q$
With $p,q,r$ belonging to $[\arccos (\frac{1}{\sqrt{2}});\arccos(\frac{1}{2})]$ :
We get this :
$$\sum_{cyc}\tan^4 s\geq u\sum_{cyc}\tan^4 p$$ 
we put :
$$u=\frac{\sum_{cyc}\tan^2 s\tan^2 v}{\sum_{cyc}\tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$u\sum_{cyc}\tan^4 p=\sum_{cyc}\tan^2 s\tan^2 v$$
And  
$$\sum_{cyc}\tan^4 s\geq\sum_{cyc}\tan^2 s\tan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2\geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution  and combine with the well-know formula $(\cos a^2+\sin a^2)^2=1$ we find:
$$\frac{\sum_{cyc}\tan^2 s\tan v^2}{\sum_{cyc}\tan^4 p}\sum_{cyc}\frac{(\tan^4 p+2\tan^2 p)((\tan^4 r+2\tan^2 r)}{\Big(1+\frac{\cos^2 q}{\cos^2 r\cos^2 p}\Big)^2}\geq\sum_{cyc}\frac{(\tan^4 s+2\tan^2 s)((\tan^4 v+2\tan^2 v)}{\Big(1+\frac{\cos^2 w}{\cos^2 v\cos^2 s}\Big)^2}$$
