# Triangle Inequality on complex numbers

### Problem

Let $$z= x + iy$$, then prove that:

$$|x| + |y| \le 2 ^{1/2} |z|$$

### Progress

I've tried to write $$|z|$$ as $$(x^2 + y^2)^{1/2}$$, and to make some algebra after this, but I'm really new at proving things, I just get to nothing.

• Well, I am really new at math, so I've tried to write |z| as (x^2 + y^2)^1/2, and to make some algebra after this, but I'm really new at proving things, I just get to nothing. Aug 29, 2014 at 2:00

Hints: 1) First square both sides.

2) Write $|z|^2=x^2+y^2$.

3) "Bring the RHS to the LHS": That is- rearrange the inequality in the form $LHS-RHS \geq 0$.

4) Now do an obvious "completing the square" argument and use the fact that the square of any real number is $\geq 0$.

• Thanks, bro. It helped a lot. I just didn't see the "completing the square". Aug 29, 2014 at 2:12
• @marco: and yet you accepted the more cryptic answer ;-). Aug 29, 2014 at 2:13
• Man, I'm just new at this, don't know what I'm really doing when clicking here. Aug 29, 2014 at 2:19
• :) no issues @marco. All the best, and happy exploring both math and this site. Aug 29, 2014 at 2:20

Hint. Since both sides are positive, the inequality is equivalent to $$(|x|+|y|)^2\le2(|x|^2+|y|^2)\ .$$ See if you can simplify this and hence prove that it is always true.

If $$z=\Re z+i\Im z$$ then not only this one but also you can prove that, $$|\Re z|,|\Im z|≤|z|$$ $$2|\Re z||\Im z|≤|z|^2$$ $$|z|≤|\Re z|+|\Im z|≤√2 |z|$$
All of these are just algebraic manipulations of $$(x-y)^2≥0;$$ $$\forall x,y \in \mathbb{R}.$$