complex conjugate; prove that $|z+w| \leq |z|+|w|$

Question

Prove that $|z+w| \leq |z|+|w|$.

Okay so I labeled $z$ as $z=a+bi$ and $w$ as $w=c+di$. Substituting and solving for everything I got $\sqrt{(a+c)^2 + (b+d)^2} \leq \sqrt{a^2 +b^2} + \sqrt{c^2 +d^2}$. I know if I plug in numbers that this statement is true but is there a way to say this in a simpler way?

you are being asked to prove complex triangle inequality. try reading this article — David L, Feb 19 '14 at 20:41
usually it helps to look at $|z|^2 = z^*z$ since square roots are annoying. — James S. Cook, Feb 19 '14 at 20:47

score 1 · Accepted Answer · answered Feb 19 '14 at 20:47

Let's try to prove that $|z+w|^2 \leq (|z| + |w|)^2$, which is an equivalent statement. The left side of the inequality is $$(z+w)(\overline{z} + \overline{w}) = |z|^2 + z \overline{w} + w \overline{z} + |w|^2.$$ On the other hand, we have $$(|z| + |w|)^2 = |z|^2 + 2|z||w| + |w|^2.$$ Try to compare the terms in the middle.

complex conjugate; prove that $|z+w| \leq |z|+|w|$

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