= Solution
For any <complex number> $w$, $\operatorname{Re}w\leq |w|$, while the <complex modulus> is multiplicative and unchanged by <complex conjugation>. Consequently
$$
|z_1+z_2|^2=|z_1|^2+|z_2|^2
+2\operatorname{Re}(z_1\overline{z_2})
\leq |z_1|^2+|z_2|^2+2|z_1||z_2|
=(|z_1|+|z_2|)^2.
$$
Both sides are nonnegative, so taking square roots proves the <triangle inequality>:
$$
\boxed{|z_1+z_2|\leq |z_1|+|z_2|.}
$$
The proposed further inequality fails for perpendicular vectors in the <complex plane>. Taking $z_1=1$, $z_2=i$ gives
$$
\boxed{|1+i|+|1-i|=2\sqrt2>2=2\max(|1|,|i|).}
$$
Back to article page
