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.

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. 
