Show that $\left| z + \sqrt{z^2 + c^2} \right| > c$ for $\operatorname{Re}(z) > 0$ I have encountered this seemingly obvious yet bizarre-to-show inequality: $$\left| z + \sqrt{z^2 + c^2} \right| > c$$ for $z \in \Bbb C$ such that $\operatorname{Re}(z) > 0$ and $c > 0$. I've tried the reverse triangular inequality but it hasn't helped. How do you show this?

I'll assume that the branch of $\sqrt{z^2 + c^2}$ is chosen such that $\operatorname{Re} \sqrt{z^2 + c^2} > 0$ (otherwise the claim is wrong).
Then
$$
\left| z + \sqrt{z^2 + c^2} \right|^2 = |z|^2 + |z^2+c^2| + 2 \operatorname{Re} \left( \bar z \cdot\sqrt{z^2 + c^2}\right) \\
\ge  |z|^2 + (c^2  - |z|^2) + 2 \operatorname{Re} \left( \bar z\cdot\sqrt{z^2 + c^2}\right) \\
 = c^2 + 2 \operatorname{Re} \left( \bar z\cdot\sqrt{z^2 + c^2}\right)
$$
so that is is sufficient to show that
$$ \tag{*}
\operatorname{Re} \left( \bar z \cdot\sqrt{z^2 + c^2}\right) > 0 \, .
$$
Now observe that if $z$ lies in the first quadrant then $\sqrt{z^2 + c^2}$ is in the first quadrant as well, and $\bar z$ is in the forth quadrant, so that the product $\bar z \cdot\sqrt{z^2 + c^2}$ is in the right half-plane. A similar argument works if $z$ lies in the forth quadrant, or on the positive real axis. In any case, $(*)$ holds.
