Prove $|(1+|u|^2)v-(1+|v|^2)u|>|u\bar{v}-\bar{u}v|$ Show that for $u,v \in \mathbb{C}$ with $|u|<1, |v|<1$, and $\bar{u}v\neq u\bar{v}$, we always have
$$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|>\left|u\bar{v}-\bar{u}v\right|.$$

I have answered here as this question which shows context has been closed as a duplicate.

Let $u=re^{i\theta}$ and $v=\rho e^{i\phi}$. Then the inequality is equivalent to $$\left|(1+r^2)\rho e^{i\phi}-(1+\rho^2)re^{i\theta}\right|>r\rho\left|e^{i(\theta-\phi)}-e^{-i(\theta-\phi)}\right|=2r\rho|\sin(\theta-\phi)|$$ and dividing both sides by $r\rho$ yields $$\left|r+\frac 1r-\left(\rho+\frac1\rho\right)e^{i(\theta-\phi)}\right|>2|\sin(\theta-\phi)|.$$ Denoting $s=r+1/r$ and $t=\rho+1/\rho$, we have $s,t>2$. Therefore, it suffices to show that $$(s-t\cos(\theta-\phi))^2+t^2\sin^2(\theta-\phi)^2>4\sin^2(\theta-\phi)$$ on squaring both sides. Denoting $\varphi=\theta-\phi$, the inequality simplifies to $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)>0.$$ When $\cos\varphi\le0$ the LHS is minimised at $s=2$ so that $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge4-4t\cos\varphi+(t^2-4+4\cos^2\varphi)=(t-2\cos\varphi)^2>0.$$ When $\cos\varphi>0$ the LHS is minimised at $s=t\cos\varphi$ so that $$s^2-2(t\cos\varphi)s+(t^2-4\sin^2\varphi)\ge-t^2\cos^2\varphi+t^2-4\sin^2\varphi=(t^2-4)\sin^2\varphi>0$$ which proves $\vert (1+\vert u \vert^2)v - (1+\vert v \vert^2)u  \vert > \vert u\overline{v}-\overline{u}v  \vert$.
