show $\sum_{cyc}(1-x)^2\ge \sum_{cyc}\frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$

Solution 1:

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:

$$\frac{\sum_{cyc}\tan^2 s\tan v^2}{\sum_{cyc}\tan^4 p}\sum_{cyc}\frac{(\tan^4 p+2\tan^2 p)((\tan^4 r+2\tan^2 r)}{\Big(1+\frac{\cos^2 q}{\cos^2 r\cos^2 p}\Big)^2}\geq\sum_{cyc}\frac{(\tan^4 s+2\tan^2 s)((\tan^4 v+2\tan^2 v)}{\Big(1+\frac{\cos^2 w}{\cos^2 v\cos^2 s}\Big)^2}$$