How to prove $|z_1+z_2|<|1+\bar{z_1}z_2|$ if $|z_1|<1$ an $|z_2|<1$ I would like to prove that $|z_1+z_2|<|1+\bar{z_1}z_2|$ if $|z_1|<1$ an $|z_2|<1$.
I tried to multiply the numerator and denominator by $|1+z_1\bar{z_2}|$ or consider the square, but I couldn't finish anything. Someone has a hint as I can solve this problem?
 A: $$\begin{aligned}\left|1+\bar{z}_{1} z_2\right|^{2} &=\left(1+\bar{z}_{1} z_{2}\right)\left(\overline{1+\bar{z}_{1} z_{2}}\right) \\ &=\left(1+\bar{z}_{1} z_{2}\right)\left(1+z_{1} \bar{z}_{2}\right)\\&=1+\bar{z}_{1} z_{2}+z_{1} \bar{z}_{2}+ \bar{z}_{1} z_{2}z_{1} \bar{z}_{2}\\& = 1+\bar{z}_{1} z_{2}+z_{1} \bar{z}_{2}+ |z_1|^{2}\left|z_{2}\right|^{2}\end{aligned}$$
$$
\begin{aligned}
\left|z_{1}+z_{2}\right|^{2} &=\left(z_{1}+z_{2}\right)\left(\overline{z_{1}+z_{2}}\right) \\
&=z_{1} \bar{z}+z_{2} \bar{z}_{1}+z_{1} \bar{z}_{2}+z_{2} \bar{z}_{2} \\
&=|z_{1} |^{2}+\bar{z_{1}} z_{2}+z_{1} \bar{z}_{2}+\left|z_{2}\right|^{2}
\end{aligned}
$$
$$
\begin{array}{l}
\therefore\left|1+\bar{z}_{1} z_{2}\right|^{2}-\left|z_{1}+z_{2}\right|^{2} \\
=1+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left.\left|z_{1}|^{2}-\right| z_{2}\right|^{2} \\
=\left(1-\mid z_1|^{2}\right)\left(1-|z_2|^{2}\right) \\
>0 \quad\left(\because\left|z_{1}\right|<1 \text { and }\left|z_{2}\right|<1\right)
\end{array}
$$
Therefore $\left|1+\bar{z}_{1} z_{2}\right|^{2}>\left|z_{1}+z_{2}\right|^{2}$ and hence $\left|1+\bar{z}_{1} z_{2}\right|>\left|z_{1}+z_{2}\right|.$
A: Denote: $z_1 = a_1+ib_1, z_2 = a_2 +ib_2\implies (a_1+a_2)^2+(b_1+b_2)^2 < (1+a_1a_2+b_1b_2)^2+(a_1b_2 - a_2b_1)^2 \iff a_1^2+ 2a_1a_2+a_2^2 + b_1^2+2b_1b_2+b_2^2 < 1 + 2a_1a_2+2b_1b_2 + (a_1^2+ b_1^2)(a_2^2+b_2^2)$. Let $x = a_1^2+b_1^2 < 1, y = a_2^2+b_2^2 < 1$, we show: $x + y < 1 + xy\iff (x - 1)(1- y) < 0$, which is clear. Thus we're done !
