prove that $|z_1-z_2| \leq |z_1| + |z_2|$

I need to prove that ,

$$|z_1 - z_2| \leq |z_1| + |z_2|$$

\begin{align} &|z_1 + z_2 |^2 \\&= (z_1+z_2) \overline {(z_1+z_2)} \\& = (z_1+z_2) (\bar{z_1} + \bar{z_2} ) \\&=(z_1\bar{z_1} + z_1\bar{z_2} + z_2\bar{z_1} + z_2\bar{z_2}) \\& =(z_1\bar{z_1} + z_2\bar{z_2} + z_1\bar{z_2} + \overline{z_1\bar{z_2}}) \\& =|z_1|^2 + |z_2|^2 + 2Re(z_1\bar{z_2}) \\& \leq |z_1|^2 + |z_2|^2 + 2|z_1\bar{z_2}| \\&= |z_1|^2 + |z_2|^2 + 2|z_1||\bar{z_2}| \\&= |z_1|^2 + |z_2|^2 + 2|z_1||{z_2}| \\&= (|z_1|+|z_2|)^2 \end{align}

Therefore , $$|z_1 + z_2 | \leq |z_1| + |z_2|$$

Now,

$$|z_1-z_2|=|z_1+(-z_2)|\leq |z_1| + |(-z_2)|$$

So ,$$|z_1-z_2| \leq |z_1| + |z_2|$$
(SOLVED) Is it correct ?If there are other ways to solve this let me know.

• Here's a tutorial in MathJax Jul 2, 2017 at 18:10
• set $$z_1=a+bi,z_2=x+iy$$ Jul 2, 2017 at 18:12
• The posted proof is correct (assuming $\operatorname{Re}(z) \le |z|$ is "obvious" enough to go without a note).
– dxiv
Jul 2, 2017 at 19:56

If you know $|z_1+z_2| \le |z_1|+|z_2|,$ then $|z_1-z_2|= |z_1+(-z_2)| \le |z_1|+|(-z_2)|= |z_1|+|z_2|.$
Geometrically, $\,z_1, z_2\,$ and $\, z_1-z_2\,$ are associated with vectors $\,\overrightarrow{OZ_1}, \overrightarrow{OZ_2}\,$ and $\,\overrightarrow{Z_2Z_1}\,$ in the complex plane, so $\,|z_1-z_2| \le |z_1|+|z_2|\,$ is equivalent to the triangle inequality for $\,\triangle OZ_1Z_2\,$: $|Z_1Z_2| \le |Z_1O|+|OZ_2|\,$.