Inequality "A la Rozenberg" Hello I want to solve this
The inequality is equivalent to this :
$$\frac{a^2}{b^2}cos(arctan(\sqrt{\frac{b}{a}}))^2+\cos(\arctan(\sqrt{\frac{c}{b}}))^2+\frac{c^2}{b^2}\cos(\arctan(\sqrt{\frac{a}{c}}))^2\geq \frac{3}{2b^2}$$
We put :
$\sqrt{\frac{b}{a}}=\frac{x+y}{1-xy}$
$\sqrt{\frac{c}{b}}=\frac{z+y}{1-yz}$
$\sqrt{\frac{a}{c}}=\frac{x+z}{1-xz}$
Where $x$,$y$,$z$ are positive real numbers with the condition 
$1>xy$,$1>xz$,$1>zy$.
We get :
$$(\frac{1-xy}{x+y})^4\cos(\arctan(\frac{x+y}{1-xy}))^2+\cos(\arctan(\frac{z+y}{1-yz}))^2+(\frac{z+y}{1-yz})^4\cos(\arctan(\frac{x+z}{1-xz}))^2\geq \frac{3^{\frac{1}{3}}}{2}(1+(\frac{1-xy}{x+y})^6+(\frac{z+y}{1-yz})^6)^{\frac{2}{3}}$$
Wich is equivalent to : 
$$(\frac{1-xy}{x+y})^4(\frac{1}{(\frac{x+y}{1-xy})^2+1})+\frac{1}{(\frac{z+y}{1-yz})^2+1}+(\frac{z+y}{1-yz})^4\frac{1}{(\frac{x+z}{1-xz})^2+1}\geq \frac{3^{\frac{1}{3}}}{2}(1+(\frac{1-xy}{x+y})^6+(\frac{z+y}{1-yz})^6)^{\frac{2}{3}}$$
And if we make a substitution like this  :
$A=\frac{1-xy}{x+y}$
$B=\frac{z+y}{1-yz}$
$C=\frac{x+z}{1-xz}$
We get :
$$A^4(\frac{1}{(\frac{1}{A})^2+1})+\frac{1}{(B)^2+1}+B^4\frac{1}{(C)^2+1}\geq \frac{3^{\frac{1}{3}}}{2}(1+(A)^6+(B)^6)^{\frac{2}{3}}$$
With the condition :
$(\frac{\frac{1}{A^6}+1+B^6}{3})^{-1}+(\frac{C^6+1+\frac{1}{B^6}}{3})^{-1}+(\frac{\frac{1}{C^6}+1+A^6}{3})^{-1}=3$
Wich is equivalent to :
$(\frac{3A^6}{A^6B^6+1+A^6})+(\frac{3B^6}{C^6B^6+1+B^6})+(\frac{3C^6}{A^6C^6+1+C^6})=3$
After that I can't continue.
Thanks.
 A: To start we study the following function :
$$f(x)=\frac{x^3}{x^2+1}+\frac{1}{h^2+1}+\frac{h}{y^2+1}-\frac{3^{\frac{2}{3}}}{2}(x^3+h^3+1)^{\frac{1}{3}}$$
The derivative of function is :
$$f'(x)=\frac{1}{2}x^2[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]$$
So we study this part of the derivative:
$$[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]$$
We remark this :
$$[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]\geq [\frac{-3^{\frac{2}{3}}}{(x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]>0$$
So the derivative is positive and the function is increasing :
So we have this :
$$f(x)\geq f(0)$$
Wich is equivalent to :
$$f(x)\geq \frac{1}{h^2+1}+\frac{h}{y^2+1}-\frac{3^{\frac{2}{3}}}{2}(h^3+1)^{\frac{1}{3}}$$
Now we combine this with the initial condition :
$$\frac{3h}{hy+1+y}+\frac{3y}{1+y}=3$$
We get :
$$\frac{1}{h^2+1}+\frac{h}{(h-1)^2+1}-\frac{3^{\frac{2}{3}}}{2}(h^3+1)^{\frac{1}{3}}\geq 0$$
Done!
