parmenides51 wrote: Let $x, y, z$ denote positive real numbers for which $x+y+z = 1$ and $x > yz$, $y > zx$, $z > xy$. Prove that $$\left(\frac{x - yz}{x + yz}\right)^2+ \left(\frac{y - zx}{y + zx}\right)^2+\left(\frac{z - xy}{z + xy}\right)^2< 1.$$ It's $$(x-yz)(y-xz)(z-xy)>0.$$

parmenides51 wrote: Let $x, y, z$ denote positive real numbers for which $x+y+z = 1$ and $x > yz$, $y > zx$, $z > xy$. Prove that $$\left(\frac{x - yz}{x + yz}\right)^2+ \left(\frac{y - zx}{y + zx}\right)^2+\left(\frac{z - xy}{z + xy}\right)^2< 1.$$ Let $A,B,C$ be the three angles of $\triangle ABC.$ From $\tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1,$ We can let $x=\tan\frac B2\tan\frac C2,y=\tan\frac C2\tan\frac A2,z=\tan\frac A2\tan\frac B2.$ From $x > yz,y > zx,z > xy,$ we get $A,B,C<\frac{\pi }{2}.$ As $\frac{x - yz}{x + yz}=\dfrac{1-\frac{yz}{x}}{1+\frac{yz}{x}}=\frac{1-\tan ^2\frac A2}{1+\tan ^2\frac A2}=\cos A,\frac{y - zx}{y + zx}=\cos B,\frac{z-xy}{z+xy}=\cos C,$ $\begin{aligned}\left(\frac{x - yz}{x + yz}\right)^2+ \left(\frac{y - zx}{y + zx}\right)^2+\left(\frac{z - xy}{z + xy}\right)^2&=\cos ^2A+\cos ^2B+\cos ^2C\\&=1-2\cos A\cos B\cos C< 1\end{aligned}$