Prove that $|z| \le |z-1|+1$ and $|z-1| \le |z|+1$ Prove that for any $ z \in \mathbb{C}$:


*

*$|z| \le |z-1|+1$

*$|z-1| \le |z|+1$


I take any $z \in \mathbb{C}$ such that $z=a+bi$ for some $a,b \in \mathbb{R}$. I received:


*

*$\sqrt{a^2 +b^2} \le \sqrt{a^2+b^2 -2a+1}+1$

*$\sqrt{a^2+b^2 -2a+1}\le \sqrt{a^2+b^2}+1$ 


But I don't know how can I prove that these inequality are always true.
 A: Both follow by triangle inequality of the norm.
(1) $|z|=|(z-1)+1|\le|z-1|+1$
(2) $|z-1|\le|z|+|-1|=|z|+1$
A: We do it the ugly way, to show it can be done. But the right way is to use the Triangle Inequality. For example, we want to show that 
$$\sqrt{a^2+b^2} \le \sqrt{(a-1)^2+b^2}+1.\tag{1}$$
(Note that we did not expand yet. Why bother, unless it is useful?)
Since both sides are non-negative, Inequality (1) is equivalent to the following inequality, obtained by squaring both sides:
$$a^2+b^2\le (a-1)^2+b^2+2\sqrt{(a-1)^2 +b^2}+1.\tag{2}$$
This inequality is equivalent to 
$$2a-2\le 2\sqrt{(a-1)^2+b^2}.$$
(Now we did expand one of the $(a-1)^2$.)
But $b^2\ge 0$, so $2\sqrt{(a-1)^2+b^2}\le 2\sqrt{(a-1)^2}=2|a-1|$. Thus we only need to show that 
$$2(a-1)\le 2|a-1|.$$
This is obvious.
A similar manipulation takes care of the second inequality. 
A: If you know the triangle inequality, these are immediate; recall that this states that for $x, y \in \mathbb{C}$,
$$|x + y| \leq |x| + |y|$$
For the first, set $x = z - 1$, $y = 1$.
The second is similar.
