Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$ if $|z_{1}| < 1$ and $|z_{2}| < 1$. Given $z_{1},z_{2} \in \mathbb{C}$
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$, if $|z_{1}| < 1$ and $|z_{2}| < 1$.
My Professor gave my classmates and I, the following answer:
$|z_{1} + z_{2}| \leq |z_{1} + z_{2}|\cdot |\overline{z_{1}}|\;$ $\Leftarrow$ The part I'm having trouble with
$|z_{1} + z_{2}|\cdot |\overline{z_{1}}| = |z_{1}\overline{z_{1}} + z_{2}\overline{z_{1}}| = ||z_{1}|^{2} + z_{2}\overline{z_{1}}| < |1+\overline{z_{1}}z_{2}|$
$\therefore |z_{1}+z_{2}| < |1+\overline{z_{1}}z_{2}|$
I don't understand why the line highlighted is correct.
Since $|\overline{z_{1}}| = |z_{1}| < 1$, $\;|\overline{z_{1}}|\cdot  |z_{1} + z_{2}|< 1\cdot |z_{1} + z_{2}|$ which contradicts the proof above.
If my reasoning is correct, How could I prove this statement?
 A: The inequality $\lvert z_1 + z_2\rvert \leq \lvert z_1 + z_2\rvert \cdot \lvert \overline{z_{1}} \rvert$ is obviously wrong. Take for example $z_1 = 0$ and $z_2 \neq 0$.
Now coming back to the initial problem, you can prove that for any reals $0 \le a,b \lt 1$, the inequality $a+b \lt 1+ab$ holds. This can be done for example by studying the real map
$$f_a(x) = ax -a -x +1$$ on the interval $[0,1]$, or by noticing that $(1-a)(1-b) \gt 0$.
Applying that to $a= \lvert z_1 \rvert^2$ and $b = \lvert z_2 \rvert^2$, you get the inequality
$$\lvert z_1 \rvert^2 + \lvert z_2 \rvert^2 \lt 1 + \lvert z_1 \rvert^2 \lvert z_2 \rvert^2$$
Adding $z_1 \overline{z_2} + z_2 \overline{z_1}$ (which is a real number) on both side of the inequality, you get the desired inequality
$$\lvert z_1 + z_2 \rvert^2 \lt \lvert 1 + \overline{z_1} z_2 \rvert^2.$$
A: Square both sides:
\begin{align*}
|z + w| < |1 + \overline{z}w| & \Longleftrightarrow |z + w|^{2} < |1 + \overline{z}w|^{2}\\\\
& \Longleftrightarrow (z + w)(\overline{z} + \overline{w}) < (1 + \overline{z}w)(1 + z\overline{w})\\\\
& \Longleftrightarrow z\overline{z} + z\overline{w} + \overline{z}w + w\overline{w} < 1 + z\overline{w} + \overline{z}w + z\overline{z}w\overline{w}\\\\
& \Longleftrightarrow |z|^{2} + |w|^{2} < 1 + |z|^{2}|w|^{2}\\\\
& \Longleftrightarrow |z|^{2}(|w|^{2} - 1) + 1 - |w|^{2} > 0\\\\
& \Longleftrightarrow (1 - |z|^{2})(1 - |w|^{2}) > 0
\end{align*}
which clearly holds based on the assumption that $|z| < 1$ and $|w| < 1$.
Hopefully this helps!
A: The professor proof is wrong as observed at the beggining of mathcounterexamples.net's answer.
Your reasoning is almost correct:
"
Since $|\overline{z_{1}}| = |z_{1}| < 1$, $\;|\overline{z_{1}}|\cdot  |z_{1} + z_{2}|\color{red}{<} 1\cdot |z_{1} + z_{2}|$ which contradicts the proof above."
Since you're multiplying by $|z_1+z_2|$ and it could be $0$ you can't assure that the strict inequality is preserved. It should be $\le$.
By the way, that's not only point where your professor went wrong:
From $\left|z_1\right|^2<1$ you cannot conclude that $\left|\left|z_1\right|^2 + \overline{z_1}z_2\right| < \left|1+\overline{z_1}z_2\right|$.
