<s>[INST] Conditional inequality Let x,y,z be positive reals with $xy+yz+zx=1$. Prove the inequality $$\sum_{cyc(x,y,z)}\frac {2x(1-x^2)}{(1+x^2)^2} \le \sum_{cyc(x,y,z)} \frac x{1+x^2}.$$ I substituted $x=tan\frac{\theta}2, y=tan\frac {\phi}2,z=tan\frac {\epsilon}2$ where $0\le\theta,\phi,\epsilon\le \pi/2$. Therefore the equality gives $tan\frac {\theta}2*tan\frac {\phi}2+tan\frac {\phi}2*tan\frac{\epsilon}2+tan\frac {\epsilon}2*tan\frac {\theta}2=1=>\frac {\theta+\phi+\epsilon}2=\frac {\pi}2=>\theta+\phi+\epsilon=\pi$. Now using trigonometric formulae we can reduce the given inequality into $$\begin{align} &=> 2sin \theta* cos\theta+2sin \phi* cos\phi+2sin\epsilon* cos\epsilon \le sin\theta+sin\phi+\sin\epsilon \\&=>sin2\theta+sin2\phi+sin2\epsilon\le sin\theta+sin\phi+sin\epsilon \\&=> 4sin \theta*sin\phi*sin \epsilon \le 4cos \frac{\theta}2*cos \frac {\phi}2*cos \frac{\epsilon}2 \\&=>8sin \frac {\theta}2*sin \frac {\phi}2*sin \frac {\epsilon}2 \le 1\end{align}$$ How do i prove this?
 [/INST] We need to prove that
$$\sum_{cyc}\frac{x}{x^2+xy+xz+yz}\geq2\sum_{cyc}\frac{x(xy+xz+yz-x^2)}{(x^2+xy+xz+yz)^2}$$ or
$$\sum_{cyc}\frac{x}{(x+y)(x+z)}\geq2\sum_{cyc}\frac{x(xy+xz+yz-x^2)}{(x+y)^2(x+z)^2}$$ or
$$\sum_{cyc}x(x+y)(x+z)(y+z)^2\geq2\sum_{cyc}x(xy+xz+yz-x^2)(y+z)^2$$ or
$$\sum_{cyc}x(3x^2-xy-xz-yz)(y+z)^2\geq0$$ or
$$\sum_{cyc}x((x-y)(3x+z)-(z-x)(2x+y))(y+z)^2\geq0$$ or
$$\sum_{cyc}(x-y)((3x^2+xz)(y+z)^2-(3y^2+yz)(x+z)^2)\geq0$$ or
$$\sum_{cyc}(x-y)^2z(5xy+3xz+3yz+z^2)\geq0.$$
Done!
</s>