Inequality on $D(0,1)$

Question

Let $u,v\in\mathbb C,|v|<1,|u|<1,\overline vu\ne\overline uv$. Show that $$|(1+|v|^2)u-(1+|u|^2)v|>|\overline vu-\overline uv|$$ I have tried using $|z|² = z \overline{z}$, which takes me to the inequality: $$(1+|v|²)²|u|²+(1+|u|²)²|v|²-2(1+|v|²)(1+|u|²)\operatorname{Re}(u\overline{v}) \\ >2|u|²|v|² -2 \operatorname{Re}((u\overline{v})²) $$ $\iff$ $$(1+|v|²)²|u|²+(1+|u|²)²|v|²+2 \operatorname{Re}((u\overline{v})²) \\ >2|u|²|v|² +2(1+|v|²)(1+|u|²)\operatorname{Re}(u\overline{v}).$$ From there i tried a polar-coordinate-approach, $u=re^{i \phi}, v = se^{i \psi}$, which gave me: $$r²+2s²r²+s⁴r²+s²+2r²s²+r⁴s²+2r²s²\cos(2(\phi - \psi))\\ >2r²s²+2rs(1+r²)(1+s²)\cos(\phi-\psi).$$ For $\cos(\phi - \psi) \leq 0$ it is trivial. But i can't figure out how the terms cancel out for positive $\cos(\phi - \psi)$ Using phyt. identity and some addition thm. i got: $$r²+2s²r²+s⁴r²+s²+2r²s²+r⁴s²+2r²s²\cos(2(\phi - \psi))\\ =r²+2s²r²+s⁴r²+s²+2r²s²+r⁴s²\\ +2r²s²\left[\cos²(\phi - \psi)-1+cos²(\phi - \psi)\right]\\ =r²+2s²r²+s⁴r²+s²+r⁴s²+4r²s²\cos²(\phi-\psi).$$ From here i don't know what to do. Any help is appreciated.