Prove: $|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}|$

Question

Prove: $$|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)*|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}|$$

Except inserting $(a+bi)$ instead of $z$ (which I think will lead me to a dead end), I really don't have a good idea how to confront this exercise, any tips or hints?

I'm not student yet and I just started learning maths from zero so I might not be familiar with all the thereoms as much as a first year math student.

In my book, it sais that it's a challenge exercise (it's not a particular book, it's just the university exercise lists for the entry test for the math facultee)

Answer 1

We have that

$$|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right|=\frac12\left| z_1+ |z_1|\frac{z_2}{|z_2|} + |z_2|\frac{z_1}{|z_1|}\right|$$

then indicating $w=z_1+z_2$ we have

$$2|w|\ge \left| w+ |z_1|\frac{z_2}{|z_2|}+ |z_2|\frac{z_1}{|z_1|}\right|=|w+v|$$

which is true, indeed the complex number

$$v=|z_1|\frac{z_2}{|z_2|}+ |z_2|\frac{z_1}{|z_1|}$$

represents the vector symmetric to vector $w$ with respect to the bisector between $z_1$ and $z_2$ and then

$$|w+v|\le 2|w|$$

Answer 2

$$ |z_1+z_2| = |\frac{z_1}{m} + \frac{z_2}{m}|m, $$ with $m=(|z_1|+|z_2|)/2$. Now we can assume $w_1$ have a larger norm than $w_2$ n the following definitions and note that we can take take $w_1 = z_1/m = (1+t)e_1$ and $w_2 = z_2/m = (1-t)e_2$, with $|e_1|=|e_2|=1$, then define $f$ by, $$ 2 f(t) = |w_1+w_2|^2 = (1+t)^2 + (1-t)^2+2(1-t^2)r = 2 + 2t^2 + 2(1-t^2)r, $$ with $r= e_1\cdot e_2$ hence $-1\leq r \leq 1$. Hence

$$ f(t) = (2+r) + t^2(1-r) $$ has a minima at $t=0$ which meas that

$$ |w_1+w_2|^2 = 2f(t) \geq 2f(0) = |e_1+e_2| = |\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}| $$ and the inequality is proven.