triangle inequality of complex numbers

Theorem.  All complex numbers $z_1$ and $z_2$ satisfy the triangle inequality

$|z_1+z_z|\leqq |z_1|+|z_2|.$(1)

Proof.

${|z_1+z_2|}^2$$\mathrm{\hspace{0.33em}}=(z_1+z_2)\overline{(z_1+z_2)}$

$\mathrm{\hspace{0.33em}}=(z_1+z_2)(\overline{z_1}+\overline{z_2})$

$\mathrm{\hspace{0.33em}}=z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}{z}_2$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+2\text{Re}(z_1\overline{z_2})$

$\mathrm{\hspace{0.33em}}\leqq {|z_1|}^2+{|z_2|}^2+2|z_1\overline{z_2}|$

$\mathrm{\hspace{0.33em}}={|z_1|}^2+{|z_2|}^2+2|z_1|\cdot |\overline{z_2}|$

$\mathrm{\hspace{0.33em}}={(|z_1|+|z_2|)}^2$

Taking then the nonnegative square root, one obtains the asserted inequality.

Remark.  Since the real numbers are complex numbers, the inequality (1) and its proof are valid also for all real numbers; however the inequality may be simplified to

$${|x+y|}^2\leqq {(x+y)}^2=x^2+2xy+y^2\leqq x^2+2|x||y|+y^2={|x|}^2+2|x||y|+{|y|}^2={(|x|+|y|)}^2.$$